7.1 Exercises

Section 7.1 Introducing Rational Functions 615

Version: Fall 2007

7.1 Exercises

In Exercises 1-14, perform each of thefollowing tasks for the given rational func-tion.

i. Set up a coordinate system on a sheetof graph paper. Label and scale eachaxis.

ii. Use geometric transformations as inExamples 10, 12, and 13 to draw thegraphs of each of the following ra-tional functions. Draw the verticaland horizontal asymptotes as dashedlines and label each with its equa-tion. You may use your calculator tocheck your solution, but you shouldbe able to draw the rational functionwithout the use of a calculator.

iii. Use set-builder notation to describethe domain and range of the givenrational function.

1. f(x) = !2/x

2. f(x) = 3/x

3. f(x) = 1/(x! 4)

4. f(x) = 1/(x+ 3)

5. f(x) = 2/(x! 5)

6. f(x) = !3/(x+ 6)

7. f(x) = 1/x! 2

8. f(x) = !1/x+ 4

9. f(x) = !2/x! 5

10. f(x) = 3/x! 5

11. f(x) = 1/(x! 2)! 3

Copyrighted material. See: http://msenux.redwoods.edu/IntAlgText/1

12. f(x) = !1/(x+ 1) + 5

13. f(x) = !2/(x! 3)! 4

14. f(x) = 3/(x+ 5)! 2

In Exercises 15-22, find all vertical as-ymptotes, if any, of the graph of the givenfunction.

15. f(x) = ! 5x+ 1 ! 3

16. f(x) = 6x+ 8 + 2

17. f(x) = ! 9x+ 2 ! 6

18. f(x) = ! 8x! 4 ! 5

19. f(x) = 2x+ 5 + 1

20. f(x) = ! 3x+ 9 + 2

21. f(x) = 7x+ 8 ! 9

22. f(x) = 6x! 5 ! 8

In Exercises 23-30, find all horizontalasymptotes, if any, of the graph of thegiven function.

23. f(x) = 5x+ 7 + 9

24. f(x) = ! 8x+ 7 ! 4

616 Chapter 7 Rational Functions

Version: Fall 2007

25. f(x) = 8x+ 5 ! 1

26. f(x) = ! 2x+ 3 + 8

27. f(x) = 7x+ 1 ! 9

28. f(x) = ! 2x! 1 + 5

29. f(x) = 5x+ 2 ! 4

30. f(x) = ! 6x! 1 ! 2

In Exercises 31-38, state the domainof the given rational function using set-builder notation.

31. f(x) = 4x+ 5 + 5

32. f(x) = ! 7x! 6 + 1

33. f(x) = 6x! 5 + 1

34. f(x) = ! 5x! 3 ! 9

35. f(x) = 1x+ 7 + 2

36. f(x) = ! 2x! 5 + 4

37. f(x) = ! 4x+ 2 + 2

38. f(x) = 2x+ 6 + 9

In Exercises 39-46, find the range ofthe given function, and express your an-swer in set notation.

39. f(x) = 2x! 3 + 8

40. f(x) = 4x! 3 + 5

41. f(x) = ! 5x! 8 ! 5

42. f(x) = ! 2x+ 1 + 6

43. f(x) = 7x+ 7 + 5

44. f(x) = ! 8x+ 3 + 9

45. f(x) = 4x+ 3 ! 2

46. f(x) = ! 5x! 4 + 9

Section 7.1 Introducing Rational Functions

Version: Fall 2007

7.1 Solutions

1. Start with the graph of y = 1/x in (a). Multiplying by 2 produces the equationy = 2/x and stretches the graph vertically by a factor of 2 as shown in (b). Multiplyingby !1 produces the equation y = !2/x and reflects the graph of y = 2/x in (b) overthe x-axis to produce the graph of y = !2/x in (c).Projecting all the points of the graph in (c) onto the x-axis provides the domain:D = {x : x "= 0}. Projecting all the points on the graph in (c) onto the y-axis producesthe range: R = {y : y "= 0}.

x10

y10

y=0

x=0

x10

y10

y=0

x=0

x10

y10

y=0

x=0(a) y = 1/x. (b) y = 2/x. (c) y = !2/x.

3. Start with the graph of y = 1/x in (a). Replacing x with x ! 4 produces theequation y = 1/(x ! 4) and slides the graph of y = 1/x four units to the right toproduce the graph of y = 1/(x! 4) in (b)Projecting all the points of the graph in (b) onto the x-axis provides the domain:D = {x : x "= 4}. Projecting all the points on the graph in (b) onto the y-axisproduces the range: R = {y : y "= 0}.

x10

y10

y=0

x=0

x10

y10

y=0

x=4(a) y = 1/x. (b) y = 1/(x ! 4).

Chapter 7 Rational Functions

Version: Fall 2007

5. Start with the graph of y = 1/x in (a). Multiply by 2 to produce the equationy = 2/x, which stretches the graph of y = 1/x vertically by a factor of 2. The graphof y = 2/x is shown in (b). Finally, replacing x with x ! 5 produces the equationy = 2/(x ! 5) and slides the graph of y = 2/x five units to the right to produce thegraph of y = 2/(x! 5) in (c).Projecting all the points of the graph in (c) onto the x-axis provides the domain:D = {x : x "= 5}. Projecting all the points on the graph in (c) onto the y-axis producesthe range: R = {y : y "= 0}.

x10

y10

y=0

x=0

x10

y10

y=0

x=0

x10

y10

y=0

x=5(a) y = 1/x. (b) y = 2/x. (c) y = 2/(x ! 5).

7. Start with the graph of y = 1/x in (a). Subtract 2 to produce the equationy = 1/x! 2. This shifts the graph downward 2 units as shown in (b).Projecting all the points of the graph in (b) onto the x-axis provides the domain:D = {x : x "= 0}. Projecting all the points on the graph in (b) onto the y-axisproduces the range: R = {y : y "= !2}.

x10

y10

y=0

x=0

x10

y10

y=!2

x=0(a) y = 1/x. (b) y = 1/x ! 2.


Version: Fall 2007

9. Start with the graph of y = 1/x in (a). Multiply by !2 to produce the equationy = !2/x. This stretches the graph of y = 1/x vertically by a factor of 2, then reflectsthe graph across the x-axis as shown in (b). Subtract 5 to produce the equqtiony = !2/x! 5. This shifts the graph of y = !2/x downward 5 units as shown in (c).Projecting all the points of the graph in (c) onto the x-axis provides the domain:D = {x : x "= 0}. Projecting all the points on the graph in (c) onto the y-axis producesthe range: R = {y : y "= !5}.

x10

y10

y=0

x=0

x10

y10

y=0

x=0

x10

y10

y=!5

x=0(a) y = 1/x. (b) y = !2/x. (c) y = !2/x ! 5.

11. Start with the graph of y = 1/x in (a). Replace x with x! 2, then subtract 3 toproduce the equation y = 1/(x! 2)! 3. This will shift the graph 2 units to the rightand 3 units downward, as shown in (b).Projecting all the points of the graph in (b) onto the x-axis provides the domain:D = {x : x "= 2}. Projecting all the points on the graph in (b) onto the y-axisproduces the range: R = {y : y "= !3}.

x10

y10

y=0

x=0

x10

y10

y=!3

x=2(a) y = 1/x. (b) y = 1/(x ! 2) ! 3.


Version: Fall 2007

13. Start with the graph of y = 1/x in (a). Multiply by !2 to produce the equationy = !2/x. This stretches the graph vertically by a factor of 2 then reflects the graphacross the x-axis, as shown in (b). Replace x with x ! 3, then subtract 4 to producethe equation y = !2/(x! 3)! 4. This will shift the graph of y = !2/x three units tothe right and 4 units downward, as shown in (c).Projecting all the points of the graph in (c) onto the x-axis provides the domain:D = {x : x "= 3}. Projecting all the points on the graph in (c) onto the y-axis producesthe range: R = {y : y "= !4}.

x10

y10

y=0

x=0

x10

y10

y=0

x=0

x10

y10

y=!4

x=3(a) y = 1/x. (b) y = !2/x. (c) y = !2/(x ! 3) ! 4.

15. The graph of f(x) = 5x+1 ! 3 can be obtained from the graph of g(x) = 1

x by (1)a shift left of 1 unit, (2) a vertical stretch by a factor of 5, (3) a reflection about thex-axis, and (4) a shift down of 3 units. Thus, the vertical asymptote x = 0 of the graphof g(x) will also shift left 1 unit to form the vertical asymptote x = !1 of the graph off(x).


x by (1)a shift left of 2 units, (2) a vertical stretch by a factor of 9, (3) a reflection about thex-axis, and (4) a shift down of 6 units. Thus, the vertical asymptote x = 0 of the graphof g(x) will also shift left 2 units to form the vertical asymptote x = !2 of the graphof f(x).

19. The graph of f(x) = 2x+5 + 1 can be obtained from the graph of g(x) = 1

x by (1)a shift left of 5 units, (2) a vertical stretch by a factor of 2, and (3) a shift up of 1 unit.Thus, the vertical asymptote x = 0 of the graph of g(x) will also shift left 5 units toform the vertical asymptote x = !5 of the graph of f(x).


x by (1)a shift left of 8 units, (2) a vertical stretch by a factor of 7, and (3) a shift down of9 units. Thus, the vertical asymptote x = 0 of the graph of g(x) will also shift left 8units to form the vertical asymptote x = !8 of the graph of f(x).


Version: Fall 2007

23. The graph of f(x) = 5x+7 +9 can be obtained from the graph of g(x) = 1

x by (1) ashift left of 7 units, (2) a vertical stretch by a factor of 5, and (3) a shift up of 9 units.Thus, the horizontal asymptote y = 0 of the graph of g(x) will also shift up 9 units toform the horizontal asymptote y = 9 of the graph of f(x).


x by (1)a shift left of 5 units, (2) a vertical stretch by a factor of 8, and (3) a shift down of 1unit. Thus, the horizontal asymptote y = 0 of the graph of g(x) will also shift down 1unit to form the horizontal asymptote y = !1 of the graph of f(x).


x by (1)a shift left of 1 unit, (2) a vertical stretch by a factor of 7, and (3) a shift down of 9units. Thus, the horizontal asymptote y = 0 of the graph of g(x) will also shift down9 units to form the horizontal asymptote y = !9 of the graph of f(x).


x by (1)a shift left of 2 units, (2) a vertical stretch by a factor of 5, and (3) a shift down of 4units. Thus, the horizontal asymptote y = 0 of the graph of g(x) will also shift down4 units to form the horizontal asymptote y = !4 of the graph of f(x).

31. An input of x = !5 would cause division by zero. Therefore, !5 is not in thedomain. All other possible inputs are valid.

33. An input of x = 5 would cause division by zero. Therefore, 5 is not in the domain.All other possible inputs are valid.



39. The graph of f(x) = 2x!3 + 8 can be obtained from the graph of g(x) = 1

x by(1) a shift right of 3 units, (2) a vertical stretch by a factor of 2, and (3) a shift upof 8 units. Thus, the horizontal asymptote y = 0 of the graph of g(x) will also shiftup 8 units to form the horizontal asymptote y = 8 of the graph of f(x). Similarly,Range(g) = {y : y "= 0} will shift vertically to form Range(f) = {y : y "= 8}. Thevalues of the function will become close to 8 as x# ±$, but never equal to 8.

41. The graph of f(x) = 5x!8 ! 5 can be obtained from the graph of g(x) = 1

x by (1)a shift right of 8 units, (2) a vertical stretch by a factor of 5, (3) a reflection about thex-axis, and (4) a shift down of 5 units. Thus, the horizontal asymptote y = 0 of thegraph of g(x) will also shift down 5 units to form the horizontal asymptote y = !5of the graph of f(x). Similarly, Range(g) = {y : y "= 0} will shift vertically to formRange(f) = {y : y "= !5}. The values of the function will become close to !5 asx# ±$, but never equal to !5.


Version: Fall 2007

43. The graph of f(x) = 7x+7 + 5 can be obtained from the graph of g(x) = 1

x by(1) a shift left of 7 units, (2) a vertical stretch by a factor of 7, and (3) a shift upof 5 units. Thus, the horizontal asymptote y = 0 of the graph of g(x) will also shiftup 5 units to form the horizontal asymptote y = 5 of the graph of f(x). Similarly,Range(g) = {y : y "= 0} will shift vertically to form Range(f) = {y : y "= 5}. Thevalues of the function will become close to 5 as x# ±$, but never equal to 5.


x by(1) a shift left of 3 units, (2) a vertical stretch by a factor of 4, and (3) a shift downof 2 units. Thus, the horizontal asymptote y = 0 of the graph of g(x) will also shiftdown 2 units to form the horizontal asymptote y = !2 of the graph of f(x). Similarly,Range(g) = {y : y "= 0} will shift vertically to form Range(f) = {y : y "= !2}. Thevalues of the function will become close to !2 as x# ±$, but never equal to !2.

Section 7.2 Reducing Rational Functions 633

Version: Fall 2007

7.2 Exercises

In Exercises 1-12, reduce each rationalnumber to lowest terms by applying thefollowing steps:

i. Prime factor both numerator and de-nominator.

ii. Cancel common prime factors.iii. Simplify the numerator and denomi-

nator of the result.

1. 14798

2. 3087245

3. 1715196

4. 22550

5. 1715441

6. 5624

7. 108189

8. 75500

9. 10028

10. 98147

11. 1125175

12. 30878575


In Exercises 13-18, reduce the given ex-pression to lowest terms. State all re-strictions.

13. x2 ! 10x+ 9

5x! 5

14. x2 ! 9x+ 20x2 ! x! 20

15. x2 ! 2x! 35x2 ! 7x

16. x2 ! 15x+ 54x2 + 7x! 8

17. x2 + 2x! 63x2 + 13x+ 42

18. x2 + 13x+ 42

9x+ 63

In Exercises 19-24, negate any two partsof the fraction, then factor (if necessary)and cancel common factors to reduce therational expression to lowest terms. Stateall restrictions.

19. x+ 2!x! 2

20. 4! xx! 4

21. 2x! 63! x

22. 3x+ 12!x! 4

23. 3x2 + 6x!x! 2


Version: Fall 2007

24. 8x! 2x2

x! 4

In Exercises 25-38, reduce each of thegiven rational expressions to lowest terms.State all restrictions.

25. x2 ! x! 2025! x2

26. x! x2

x2 ! 3x+ 2

27. x2 + 3x! 28x2 + 5x! 36

28. x2 + 10x+ 9x2 + 15x+ 54

29. x2 ! x! 568x! x2

30. x2 ! 7x+ 105x! x2

31. x2 + 13x+ 42x2 ! 2x! 63

32. x2 ! 16x2 ! x! 12

33. x2 ! 9x+ 14

49! x2

34. x2 + 7x+ 12

9! x2

35. x2 ! 3x! 18x2 ! 6x+ 5

36. x2 + 5x! 6x2 ! 1

37. x2 ! 3x! 10!9x! 18

38. x2 ! 6x+ 816! x2

In Exercises 39-42, reduce each ratio-nal function to lowest terms, and thenperform each of the following tasks.

i. Load the original rational expressioninto Y1 and the reduced rational ex-pression (your answer) into Y2 of yourgraphing calculator.

ii. In TABLE SETUP, set TblStart equalto zero, !Tbl equal to 1, then makesure both independent and dependentvariables are set to Auto. Select TA-BLE and scroll with the up- and down-arrows on your calculator until thesmallest restriction is in view. Copyboth columns of the table onto yourhomework paper, showing the agree-ment between Y1 and Y2 and whathappens at all restrictions.

39. x2 ! 8x+ 7x2 ! 11x+ 28

40. x2 ! 5xx2 ! 9x

41. 8x! x2

x2 ! x! 56

42. x2 + 13x+ 40!2x! 16

Given f(x) = 2x + 5, simplify each ofthe expressions in Exercises 43-46. Besure to reduce your answer to lowest termsand state any restrictions.

43. f(x)! f(3)x! 3

44. f(x)! f(6)x! 6

45. f(x)! f(a)x! a

46. f(a+ h)! f(a)h

Section 7.2 Reducing Rational Functions 635

Version: Fall 2007

Given f(x) = x2 + 2x, simplify each ofthe expressions in Exercises 47-50. Besure to reduce your answer to lowest termsand state any restrictions.

47. f(x)! f(1)x! 1

48. f(x)! f(a)x! a

49. f(a+ h)! f(a)h

50. f(x+ h)! f(x)h

Drill for Skill. In Exercises 51-54,evaluate the given function at the givenexpression and simplify your answer.

51. Suppose that f is the function

f(x) = ! x! 68x+ 7 .

Evaluate f(!3x + 2) and simplify youranswer.


f(x) = !5x+ 37x+ 6 .

Evaluate f(!5x + 1) and simplify youranswer.


f(x) = !3x! 64x+ 6 .

Evaluate f(!x!3) and simplify your an-swer.


f(x) = 4x! 12x! 4 .

Evaluate f(5x) and simplify your answer.


Version: Fall 2007

7.2 Solutions

1.14798 = 3 · 7 · 7

2 · 7 · 7 = 32 = 3

2

3.1715196 = 5 · 7 · 7 · 7

2 · 2 · 7 · 7 = 5 · 72 · 2 = 35

4

5.1715441 = 5 · 7 · 7 · 7

3 · 3 · 7 · 7 = 5 · 73 · 3 = 35

9

7.108189 = 2 · 2 · 3 · 3 · 3

3 · 3 · 3 · 7 = 2 · 27 = 4

7

9.10028 = 2 · 2 · 5 · 5

2 · 2 · 7 = 5 · 57 = 25

7

11.1125175 = 3 · 3 · 5 · 5 · 5

5 · 5 · 7 = 3 · 3 · 57 = 45

7

13.x2 ! 10x+ 9

5x! 5 = (x! 1)(x! 9)5(x! 1) = x! 9

5 ,

provided x "= 1

15.x2 ! 2x! 35x2 ! 7x = (x! 7)(x+ 5)

x(x! 7) = x+ 5x,

provided x "= 0, 7

Section 7.2 Reducing Rational Functions

Version: Fall 2007

17.x2 + 2x! 63x2 + 13x+ 42 = (x! 7)(x+ 9)

(x+ 7)(x+ 6) ,

provided x "= !7, !6, and this expression does not simplify any further.

19. Negate the denominator and the fraction bar.x+ 2!x! 2 = !x+ 2

x+ 2 = !1

This result is valid provided x "= !2.

21. Negate the denominator and the fraction bar.2x! 63! x = !2x! 6

x! 3Factor, then cancel.

!2x! 6x! 3 = !2(x! 3)

x! 3 = !2(x! 3)x! 3 = !2

This result is valid provided x "= 3.

23. Negate the denominator and the fraction bar.3x2 + 6x!x! 2 = !3x2 + 6x

x+ 2Factor, then cancel.

!3x2 + 6xx+ 2 = !3x(x+ 2)

x+ 2 = !3x(x+ 2)x+ 2 = !3x (1)

This result is valid provided x "= !2.

25.x2 ! x! 20

25! x2 = !x2 ! x! 20x2 ! 25 = !(x! 5)(x+ 4)

(x! 5)(x+ 5) = !x+ 4x+ 5 ,

provided x "= !5, 5

27.x2 + 3x! 28x2 + 5x! 36 = (x! 4)(x+ 7)

(x! 4)(x+ 9) = x+ 7x+ 9 ,

provided x "= 4, !9


Version: Fall 2007

29.x2 ! x! 56

8x! x2 = !x2 ! x! 56x2 ! 8x = !(x! 8)(x+ 7)

x(x! 8) = !x+ 7x,

provided x "= 0, 8

31.x2 + 13x+ 42x2 ! 2x! 63 = (x+ 7)(x+ 6)

(x+ 7)(x! 9) = x+ 6x! 9 ,

provided x "= !7, 9

33.x2 ! 9x+ 14

49! x2 = !x2 ! 9x+ 14x2 ! 49 = !(x! 7)(x! 2)

(x! 7)(x+ 7) = !x! 2x+ 7 ,

provided x "= 7, !7

35.x2 ! 3x! 18x2 ! 6x+ 5 = (x! 6)(x+ 3)

(x! 1)(x! 5) ,

provided x "= 1, 5, and this expression does not simplify any further.

37.x2 ! 3x! 10!9x! 18 = (x+ 2)(x! 5)

!9(x+ 2) = !x! 59

provided x "= !2

39.x2 ! 8x+ 7x2 ! 11x+ 28 = (x! 7)(x! 1)

(x! 7)(x! 4) = x! 1x! 4 ,

provided x "= 7, 4

X Y1 Y23 -2 -24 Err: Err:5 4 46 2.5 2.57 Err: 28 1.75 1.75

Section 7.2 Reducing Rational Functions

Version: Fall 2007

41.8x! x2

x2 ! x! 56 = ! x2 ! 8xx2 ! x! 56 = ! x(x! 8)

(x! 8)(x+ 7) = ! x

x+ 7 ,

provided x "= !7, 8

X Y1 Y2-8 -8 -8-7 Err: Err:-6 6 6-5 2.5 2.5-4 1.33333 1.33333-3 0.75 0.75-2 0.4 0.4-1 0.166667 0.1666670 -0 -01 -0.125 -0.1252 -0.222222 -0.2222223 -0.3 -0.34 -0.363636 -0.3636365 -0.416667 -0.4166676 -0.461538 -0.4615387 -0.5 -0.58 Err: -0.5333339 -0.5625 -0.5625

43.f(x)! f(3)x! 3 = (2x+ 5)! (2(3) + 5)

x! 3= 2x! 6x! 3

= 2(x! 3)x! 3

= 2,

provided x "= 3.


Version: Fall 2007

45.f(x)! f(a)x! a = (2x+ 5)! (2(a) + 5)

x! 3= 2x! 2ax! a

= 2(x! a)x! a

= 2,

provided x "= a.

47.f(x)! f(1)x! 1 = (x2 + 2x)! ((1)2 + 2(1))

x! 1

= x2 + 2x! 3x! 1

= (x+ 3)(x! 1)x! 1

= x+ 3,

provided x "= 1.

49.f(a+ h)! f(a)

h= [(a+ h)2 + 2(a+ h)]! [a2 + 2a]

h

= a2 + 2ah+ h2 + 2a+ 2h! a2 ! 2a

h

= 2ah+ h2 + 2hh

= h(2a+ h+ 2)h

= 2a+ h+ 2,

provided h "= 0.

51. Substitute !3x+ 2 for x in ! x!68x+7 and simplify to get ! 3x+4

24x!23 .

53. Substitute !x! 3 for x in !3x!64x+6 and simplify to get !3x+15

4x+6 .

Section 7.3 Graphing Rational Functions 655

Version: Fall 2007

7.3 Exercises

For rational functions Exercises 1-20,follow the Procedure for Graphing Ratio-nal Functions in the narrative, perform-ing each of the following tasks.For rational functions Exercises 1-20,perform each of the following tasks.

i. Set up a coordinate system on graphpaper. Label and scale each axis. Re-member to draw all lines with a ruler.

ii. Perform each of the nine steps listedin the Procedure for Graphing Ratio-nal Functions in the narrative.

1. f(x) = (x! 3)/(x+ 2)

2. f(x) = (x+ 2)/(x! 4)

3. f(x) = (5! x)/(x+ 1)

4. f(x) = (x+ 2)/(4! x)

5. f(x) = (2x! 5)/(x+ 1)

6. f(x) = (2x+ 5/(3! x)

7. f(x) = (x+ 2)/(x2 ! 2x! 3)

8. f(x) = (x! 3)/(x2 ! 3x! 4)

9. f(x) = (x+ 1)/(x2 + x! 2)

10. f(x) = (x! 1)/(x2 ! x! 2)

11. f(x) = (x2 ! 2x)/(x2 + x! 2)

12. f(x) = (x2 ! 2x)/(x2 ! 2x! 8)

13. f(x) = (2x2!2x!4)/(x2!x!12)

14. f(x) = (8x! 2x2)/(x2 ! x! 6)


15. f(x) = (x! 3)/(x2 ! 5x+ 6)

16. f(x) = (2x! 4)/(x2 ! x! 2)

17. f(x) = (2x2 ! x! 6)/(x2 ! 2x)

18. f(x) = (2x2 ! x! 6)/(x2 ! 2x)

19. f(x) = (4+2x!2x2)/(x2 +4x+3)

20. f(x) = (3x2 ! 6x! 9)/(1! x2)

In Exercises 21-28, find the coordinatesof the x-intercept(s) of the graph of thegiven rational function.

21. f(x) = 81! x2

x2 + 10x+ 9

22. f(x) = x! x2

x2 + 5x! 6

23. f(x) = x2 ! x! 12x2 + 2x! 3

24. f(x) = x2 ! 81x2 ! 4x! 45

25. f(x) = 6x! 18x2 ! 7x+ 12

26. f(x) = 4x+ 36x2 + 15x+ 54

27. f(x) = x2 ! 9x+ 14x2 ! 2x

28. f(x) = x2 ! 5x! 36x2 ! 9x+ 20


Version: Fall 2007

In Exercises 29-36, find the equationsof all vertical asymptotes.

29. f(x) = x2 ! 7xx2 ! 2x

30. f(x) = x2 + 4x! 453x+ 27

31. f(x) = x2 ! 6x+ 8x2 ! 16

32. f(x) = x2 ! 11x+ 18

2x! x2

33. f(x) = x2 + x! 12!4x+ 12

34. f(x) = x2 ! 3x! 549x! x2

35. f(x) = 16! x2

x2 + 7x+ 12

36. f(x) = x2 ! 11x+ 30!8x+ 48

In Exercises 37-42, use a graphing cal-culator to determine the behavior of thegiven rational function as x approachesboth positive and negative infinity by per-forming the following tasks:

i. Load the rational function into the Y=menu of your calculator.

ii. Use the TABLE feature of your calcula-tor to determine the value of f(x) forx = 10, 100, 1000, and 10000. Recordthese results on your homework in ta-ble form.

iii. Use the TABLE feature of your calcula-tor to determine the value of f(x) forx = !10, !100, !1000, and !10000.Record these results on your home-work in table form.

iv. Use the results of your tabular explo-ration to determine the equation of

the horizontal asymptote.

37. f(x) = (2x+ 3)/(x! 8)

38. f(x) = (4! 3x)/(x+ 2)

39. f(x) = (4! x2)/(x2 + 4x+ 3)

40. f(x) = (10! 2x2)/(x2 ! 4)

41. f(x) = (x2!2x!3)/(2x2!3x!2)

42. f(x) = (2x2 ! 3x! 5)/(x2 ! x! 6)

In Exercises 43-48, use a purely ana-lytical method to determine the domainof the given rational function. Describethe domain using set-builder notation.

43. f(x) = x2 ! 5x! 6!9x! 9

44. f(x) = x2 + 4x+ 3x2 ! 5x! 6

45. f(x) = x2 + 5x! 24x2 ! 3x

46. f(x) = x2 ! 3x! 4x2 ! 5x! 6

47. f(x) = x2 ! 4x+ 3x! x2

48. f(x) = x2 ! 4x2 ! 9x+ 14

Section 7.3 Graphing Rational Functions

Version: Fall 2007

7.3 Solutions

1. Step 1: The numerator and denominator of

f(x) = x! 3x+ 2

are already factored.Step 2: Note that x = !2 makes the denominator zero and is a restriction.Step 3: The number x = 3 will make the numerator of the rational function f(x) =(x! 3)/(x+ 2) equal to zero without making the denominator equal to zero (it is not arestriction). Hence, x = 3 is a zero and (3, 0) will be an x-intercept of the graph of f .Step 4: Since f is already reduced to lowest terms, we note that the restriction x = !2will place a vertical asymptote in the graph of f with equation x = !2.Step 5: We will calculate and plot two points, one on each side of the vertical asymp-tote: (!1,!4) and (!3, 6).Step 6: To find the horizontal asymptote, we use our calculator (see images (a), (b),and (c) below) to determine the end-behavior of f .

(a) (b) (c)

Thus, the line y = 1 is a horizontal asymptote.Step 7: Putting all of this information together allows us to draw the following graph.

x10

y10

x=!2

y=1

(3,0)

Step 8: The restrictions of the reduced form are the same as the restrictions of theoriginal form. Hence, there are no “holes” in the graph.


Version: Fall 2007

Step 9: We can use our graphing calculator to check our graph, as shown in thefollowing sequence of images.

(a) (b) (c)


f(x) = 5! xx+ 1

are already factored.Step 2: Note that x = !1 makes the denominator zero and is a restriction.Step 3: Note that the number x = 5 will make the numerator of the rational functionf(x) = (5 ! x)/(x + 1) equal to zero without making the denominator equal to zero(it is not a restriction). Hence, x = 5 is a zero and (5, 0) will be an x-intercept of thegraph of f .Step 4: Since f is already reduced to lowest terms, we note that the restriction x = !1will place a vertical asymptote in the graph of f with equation x = !1.Step 5: We will calculate and plot two points, one on each side of the vertical asymp-tote: (!2,!7) and (0, 5).Step 6: To find the horizontal asymptote, we use our calculator (see images (a), (b),and (c) below) to determine the end-behavior of f .

(a) (b) (c)

Thus, the line y = !1 is a horizontal asymptote.Step 7: Putting all of this information together allows us to draw the following graph.


Version: Fall 2007

x10

y10

x=!1

y=!1

(5,0)

Step 8: The restrictions of the reduced form are the same as the restrictions of theoriginal form. Hence, there are no “holes” in the graph.Step 9: We can use our graphing calculator to check our graph, as shown in thefollowing sequence of images.

(a) (b) (c)


f(x) = 2x! 5x+ 1

are already factored.Step 2: Note that x = !1 makes the denominator zero and is a restriction.Step 3: Note that the number x = 5/2 will make the numerator of the rational functionf(x) = (2x ! 5)/(x + 1) equal to zero without making the denominator equal to zero(it is not a restriction). Hence, x = 5/2 is a zero and (5/2, 0) will be an x-intercept ofthe graph of f .Step 4: Since f is already reduced to lowest terms, we note that the restriction x = !1will place a vertical asymptote in the graph of f with equation x = !1.Step 5: We will calculate and plot two points, one on each side of the vertical asymp-tote: (!2, 9) and (0,!5).Step 6: To find the horizontal asymptote, we use our calculator (see images (a), (b),and (c) below) to determine the end-behavior of f .


Version: Fall 2007

(a) (b) (c)


x10

y10

x=!1

y=2

(5/2,0)(5/2,0)


(a) (b) (c)

7. Step 1: Factor numerator and denominator to obtain

f(x) = x+ 2(x+ 1)(x! 3) .

Step 2: Note the restrictions are x = !1 and x = 3.Step 3: The number x = !2 will make the numerator of the rational function f(x) =(x + 2)/((x + 1)(x ! 3)) equal to zero without making the denominator equal to zero


Version: Fall 2007

(it is not a restriction). Hence, x = !2 is a zero and (!2, 0) will be an x-intercept ofthe graph of f .Step 4: Since f is already reduced to lowest terms, we note that the restrictions x = !1and x = 3 will place vertical asymptotes in the graph of f with equations x = !1 andx = 3.Step 5: We will calculate and plot points, one on each side of the vertical asymptote:(!1.5, 0.2222), (0,!0.6667), and (4, 1.2). These points are approximations, not exact.Step 6: To find the horizontal asymptote, we use our calculator (see images (a), (b),and (c) below) to determine the end-behavior of f .

(a) (b) (c)


x10

y10

(!2,0)

x=3x=!1

y=0

Step 8: The restrictions of the reduced form are the same as the restrictions of theoriginal form. Hence, there are no “holes” in the graph.Step 9: We can use our graphing calculator to check our graph, as shown in thefollowing sequence of images. Note that the image in (c) is not very good, but it isenough to convince us that our work above is reasonable.

(a) (b) (c)


Version: Fall 2007


f(x) = x+ 1(x+ 2)(x! 1) .

Step 2: The restrictions are x = !2 and x = 1.Step 3: The number x = !1 will make the numerator of the rational function f(x) =(x + 1)/((x + 2)(x ! 1)) equal to zero without making the denominator equal to zero(it is not a restriction). Hence, x = !1 is a zero and (!1, 0) will be an x-intercept ofthe graph of f .Step 4: Since f is already reduced to lowest terms, we note that the restrictions x = !2and x = 1 will place vertical asymptotes in the graph of f with equations x = !2 andx = 1.Step 5: We will calculate and plot points, one on each side of the vertical asymptote:(!3,!0.5), (0,!0.5), and (2, 0.75). These points are approximations, not exact.Step 6: To find the horizontal asymptote, we use our calculator (see images (a), (b),and (c) below) to determine the end-behavior of f .

(a) (b) (c)

Thus, a horizontal asymptote is the line y = 0.Step 7: Putting all of this information together allows us to draw the following graph.

x5

y5

x=!2 x=1

y=0 (!1,0)(!1,0)

Step 8: The restrictions of the reduced form are the same as the restrictions of theoriginal form. Hence, there are no “holes” in the graph.


Version: Fall 2007

Step 9: We can use our graphing calculator to check our graph, as shown in thefollowing sequence of images. Note that the image in (c) is not very good, but it isenough to convince us that our work above is reasonable.

(a) (b) (c)


f(x) = x(x! 2)(x+ 2)(x! 1) .

Step 2: The restrictions are x = !2 and x = 1.Step 3: The numbers x = 0 and x = 2 will make the numerator of the rational functionf(x) = x(x! 2)/((x+ 2)(x! 1)) equal to zero without making the denominator equalto zero (they are not restrictions). Hence, x = 0 and x = 2 are zeros and (0, 0) and(2, 0) will be x-intercepts of the graph of f .Step 4: Since f is already reduced to lowest terms, we note that the restrictions x = !2and x = 1 will place vertical asymptotes in the graph of f with equations x = !2 andx = 1.Step 5: We will calculate and plot points, one on each side of the vertical asymptote:(!3, 3.75), (!1,!1.5), (0.5, 0.6), and (1.5,!0.4286). These points are approximations,not exact.Step 6: To find the horizontal asymptote, we use our calculator (see images (a), (b),and (c) below) to determine the end-behavior of f .

(a) (b) (c)



Version: Fall 2007

x10

y10

x=!2 x=1

y=1 (0,0)(0,0)(2,0)(2,0)

Step 8: The restrictions of the reduced form are the same as the restrictions of theoriginal form. Hence, there are no “holes” in the graph.Step 9: We can use our graphing calculator to check our graph, as shown in thefollowing sequence of images. Note that the image in (c) is not very good, but it isenough to convince us that our work above is reasonable.

(a) (b) (c)

13. Step 1: Factor.

f(x) = 2x2 ! 2x! 4x2 ! x! 12 = 2(x2 ! x! 2)

x2 ! x! 12 = 2(x! 2)(x+ 1)(x! 4)(x+ 3)

Step 2: The restrictions are x = 4 and x = !3.Step 3: The numbers x = 2 and x = !1 will make the numerator of the rationalfunction f equal to zero without making the denominator equal to zero (they are notrestrictions). Hence, x = 2 and x = !1 are zeros and (2, 0) and (!1, 0) will be x-intercepts of the graph of f .Step 4: Since f is already reduced to lowest terms, we note that the restrictions x = 4and x = !3 will place vertical asymptotes in the graph of f with equations x = 4 andx = !3.Step 5: We will calculate and plot points, one on each side of the vertical asymptote:(!4, 4.5), (!2,!1.3333), (3,!1.3333), and (5, 4.5). These points are approximations,not exact.Step 6: To find the horizontal asymptote, we use our calculator (see images (a), (b),and (c) below) to determine the end-behavior of f .


Version: Fall 2007

(a) (b) (c)


x10

y10

x=!3 x=4

y=2 (!1,0)(!1,0) (2,0)(2,0)


(a) (b) (c)

15. Step 1: Factor.

f(x) = x! 3x2 ! 5x+ 6 = x! 3

(x! 3)(x! 2)

Step 2: The restrictions are x = 3 and x = 2.Step 3: There is no number that will make the numerator zero without making thedenominator zero. Hence, the function has no zeros and the graph has no x-intercept.Step 4: Reduce, obtaining the new function


Version: Fall 2007

g(x) = 1x! 2 .

Note that x = 2 is still a restriction of the reduced form, so the graph of f must havevertical asymptote with equation x = 2. Note that x = 3 is no longer a restriction ofthe reduced form and will cause a “hole” in the graph, which we will deal with shortly.Step 5: We will calculate and plot points, one on each side of the vertical asymptote:(1,!1) and (4, 0.5).Step 6: To find the horizontal asymptote, we use our calculator (see images (a), (b),and (c) below) to determine the end-behavior of f .

(a) (b) (c)


x5

y5

x=2

y=0(3,1)

Step 8: Recall that we determined that the graph of f will have a “hole” at therestriction x = 3. Use the reduced form g(x) = 1/(x!2) to compute the y-value of the“hole.”

g(3) = 13! 2 = 1

Hence, there will be a “hole” in the graph of f at (3, 1).Step 9: We can use our graphing calculator to check our graph, as shown in thefollowing sequence of images. Note that the image in (c) is not very good, but it isenough to convince us that our work above is reasonable.


Version: Fall 2007

(a) (b) (c)

17. Step 1: Factor,

f(x) = 2x2 ! x! 6x2 ! 2x = (2x+ 3)(x! 2)

x(x! 2)

Step 2: The restrictions are x = 0 and x = 2.Step 3: The number x = !3/2 makes the numerator zero without making the denom-inator zero (it is not a restriction). Hence, x = !3/2 is a zero of f and (!3/2, 0) is anx-intercept of the graph of f .Step 4: Reduce to obtain the new function

g(x) = 2x+ 3x

Note that x = 0 is still a restriction of the reduced form, so the graph of f must havevertical asymptote with equation x = 0. Note that x = 2 is no longer a restriction ofthe reduced form. Hence the graph of f will have a “hole” at this restriction. We willdeal with this point in a moment.Step 5: We will calculate and plot points, one on each side of the vertical asymptote:(!1,!1) and (1, 5).Step 6: To find the horizontal asymptote, we use our calculator (see images (a), (b),and (c) below) to determine the end-behavior of f .

(a) (b) (c)



Version: Fall 2007

x10

y10

x=0

y=2

(!3/2,0)(!3/2,0)

(2,7/2)

Step 8: Recall that we determined that the graph of f will have a “hole” at therestriction x = 2. Use the reduced form g(x) = (2x+ 3)/x to determine the y-value ofthe “hole.”

g(2) = 2(2) + 32 = 7

2Hence, there will be a “hole” in the graph of f at (2, 7/2).Step 9: We can use our graphing calculator to check our graph, as shown in thefollowing sequence of images.

(a) (b) (c)

19. Step 1: Factor.

f(x) = 4 + 2x! 2x2

x2 + 4x+ 3 = !2(x2 ! x! 2)x2 + 4x+ 3 = !2(x! 2)(x+ 1)

(x+ 3)(x+ 1)Step 2: The restrictions are x = !3 and x = !1.Step 3: The number x = 2 makes the numerator zero without making the denominatorzero. Hence, x = 2 is a zero of f and (2, 0) is an x-intercept of the graph of f .Step 4: Reduce, obtaining the new function

g(x) = !2(x! 2)x+ 3 .

Note that x = !3 is still a restriction of the reduced form, so the graph of f must havevertical asymptote with equation x = !3. Note that x = !1 is no longer a restrictionof the reduced form, so the graph of f will have a “hole” at this restriction. We willdeal with the coordinates of this point in a moment.


Version: Fall 2007

Step 5: We will calculate and plot points, one on each side of the vertical asymptote:(!4,!12) and (!2, 8).Step 6: To find the horizontal asymptote, we use our calculator (see images (a), (b),and (c) below) to determine the end-behavior of f .

(a) (b) (c)

Thus, the line y = !2 is a horizontal asymptote.Step 7: Putting all of this information together allows us to draw the following graph.

x10

y10

x=!3

y=!2

(2,0)(2,0)

(!1,3)

Step 8: Recall that the graph of f will have a “hole” at the restriction x = !1. Usethe reduced form g(x) = (!2(x! 2))/(x+ 3) to compute the y-value of the hole.

g(!1) = !2(!1! 2)!1 + 3 = 3

Hence, there will be a “hole” in the graph of f at (!1, 3).Step 9: We can use our graphing calculator to check our graph, as shown in thefollowing sequence of images.

(a) (b) (c)


Version: Fall 2007

21. x-intercepts occur at values where the numerator of the function is zero, but thedenominator is not zero.

81! x2

x2 + 10x+ 9 = ! x2 ! 81x2 + 10x+ 9 = !(x+ 9)(x! 9)

(x+ 9)(x+ 1)

The numerator is zero at x = !9 and x = 9, and the denominator is zero at x = !9and x = !1. Therefore, the only x-intercept occurs at x = 9.


x2 ! x! 12x2 + 2x! 3 = (x+ 3)(x! 4)

(x+ 3)(x! 1)

The numerator is zero at x = !3 and x = 4, and the denominator is zero at x = !3and x = 1. Therefore, the only x-intercept occurs at x = 4.


6x! 18x2 ! 7x+ 12 = 6(x! 3)

(x! 3)(x! 4)

The numerator is zero at x = 3, and the denominator is zero at x = 3 and x = 4.Therefore, there are no x-intercepts.


x2 ! 9x+ 14x2 ! 2x = (x! 2)(x! 7)

x(x! 2)

The numerator is zero at x = 2 and x = 7, and the denominator is zero at x = 2 andx = 0. Therefore, the only x-intercept occurs at x = 7.

29. Vertical asymptotes occur where the simplified function is not defined.x2 ! 7xx2 ! 2x = x(x! 7)

x(x! 2) = x! 7x! 2

so the line x = 2 is the only vertical asymptote.

31. Vertical asymptotes occur where the simplified function is not defined.x2 ! 6x+ 8x2 ! 16 = (x! 4)(x! 2)

(x! 4)(x+ 4) = x! 2x+ 4

so the line x = !4 is the only vertical asymptote.


Version: Fall 2007

33. Vertical asymptotes occur where the simplified function is not defined.x2 + x! 12!4x+ 12 = (x! 3)(x+ 4)

!4(x! 3) = !x+ 44

so there are no vertical asymptotes.

35. Vertical asymptotes occur where the simplified function is not defined.16! x2

x2 + 7x+ 12 = ! x2 ! 16x2 + 7x+ 12 = !(x+ 4)(x! 4)

(x+ 4)(x+ 3) = !x! 4x+ 3

so the line x = !3 is the only vertical asymptote.

37. Load the equation in (a), then determine the end-behavior in (b) and (c).

(a) (b) (c)

Hence, the equation of the horizontal asymptote is y = 2.


(a) (b) (c)

Hence, the equation of the horizontal asymptote is y = !1.


(a) (b) (c)


Version: Fall 2007

Hence, the equation of the horizontal asymptote is y = 1/2.

43. The denominator factors as !9(x+1), so an input of x = !1 would cause divisionby zero. Therefore, !1 is not in the domain. All other possible inputs are valid.

45. The denominator factors as x(x! 3), so an input of x = 3 or x = 0 would causedivision by zero. Therefore, 3 and 0 are not in the domain. All other possible inputsare valid.

47. The denominator factors as x(1 ! x), so inputs of x = 0 or x = 1 would causedivision by zero. Therefore, 0 and 1 are not in the domain. All other possible inputsare valid.

Section 7.4 Products and Quotients of Rational Functions 671

Version: Fall 2007

7.4 Exercises

In Exercises 1-10, reduce the productto a single fraction in lowest terms.

1. 10814 ·

6100

2. 7563 ·

1845

3. 18956 ·

1227

4. 4572 ·

6364

5. 1536 ·

28100

6. 18949 ·

3225

7. 21100 ·

12516

8. 2135 ·

4945

9. 5620 ·

9832

10. 27125 ·

412

In Exercises 11-34, multiply and sim-plify. State all restrictions.

11.x+ 6

x2 + 16x+ 63 ·x2 + 7xx+ 4

12.x2 + 9xx2 ! 25 ·

x2 ! x! 20!18! 11x! x2


13.x2 + 7x+ 10x2 ! 1 · !9 + 10x! x2

x2 + 9x+ 20

14.x2 + 5xx! 4 ·

x! 2x2 + 6x+ 5

15.x2 ! 5xx2 + 2x! 48 ·

x2 + 11x+ 24x2 ! x

16.x2 ! 6x! 27x2 + 10x+ 24 ·

x2 + 13x+ 42x2 ! 11x+ 18

17.!x! x2

x2 ! 9x+ 8 ·x2 ! 4x+ 3x2 + 4x+ 3

18.x2 ! 12x+ 35x2 + 2x! 15 ·

45 + 4x! x2

x2 + x! 30

19.x+ 27 ! x ·

x2 + x! 56x2 + 7x+ 6

20.x2 ! 2x! 15x2 + x · x2 + 7x

x2 + 12x+ 27

21.x2 ! 9

x2 ! 4x! 45 ·x! 6!3! x


Version: Fall 2007

22.x2 ! 12x+ 27x! 4 · x! 5

x2 ! 18x+ 81

23.x+ 5

x2 + 12x+ 32 ·x2 ! 2x! 24x+ 7

24.x2 ! 36

x2 + 11x+ 24 ·!8! xx+ 4

25.x! 5

x2 ! 8x+ 12 ·x2 ! 12x+ 36x! 8

26.x2 ! 5x! 36x! 1 · x! 5

x2 ! 81

27.x2 + 2x! 15x2 ! 10x+ 16 ·

x2 ! 7x+ 103x2 + 13x! 10

28.5x2 + 14x! 3x+ 9 · x! 7

x2 + 10x+ 21

29.x2 ! 4

x2 + 2x! 63 ·x2 + 6x! 27x2 ! 6x! 16

30.x2 + 5x+ 6x2 ! 3x · x2 ! 5x

x2 + 9x+ 18

31.x! 1

x2 + 2x! 63 ·x2 ! 81x+ 4

32.x2 + 9xx2 + 7x+ 12 ·

27 + 6x! x2

x2 ! 5x

33.5! xx+ 3 ·

x2 + 3x! 182x2 ! 7x! 15

34.4x2 + 21x+ 518! 7x! x2 ·

x2 + 11x+ 18x2 ! 25

In Exercises 35-58, divide and simplify.State all restrictions.

35.x2 ! 14x+ 48x2 + 10x+ 16!24 + 11x! x2

x2 ! x! 72

36.x! 1

x2 ! 14x+ 48 ÷x+ 5

x2 ! 3x! 18

37.x2 ! 1

x2 ! 7x+ 12 ÷x2 + 6x+ 5!24 + 10x! x2

38.x2 ! 13x+ 42x2 ! 2x! 63 ÷

x2 ! x! 42x2 + 8x+ 7

39.x2 ! 25x+ 1 ÷

5x2 + 23x! 10x! 3


Version: Fall 2007

40.x2 ! 3xx2 ! 7x+ 6x2 ! 4x

3x2 ! 11x! 42

41.x2 + 10x+ 21x! 4x2 + 3xx+ 8

42.x2 + 8x+ 15x2 ! 14x+ 45 ÷

x2 + 11x+ 30!30 + 11x! x2

43.x2 ! 6x! 16x2 + x! 42x2 ! 64

x2 + 12x+ 35

44.x2 + 3x+ 2x2 ! 9x+ 18x2 + 7x+ 6x2 ! 6x

45.x2 + 12x+ 35x+ 4

x2 + 10x+ 25x+ 9

46.x2 ! 8x+ 7x2 + 3x! 18 ÷

x2 ! 7xx2 + 6x! 27

47.x2 + x! 30x2 + 5x! 36 ÷

!6! xx+ 8

48.2x! x2

x2 ! 15x+ 54x2 + x

x2 ! 11x+ 30

49.x2 ! 9x+ 8x2 ! 9x2 ! 8x

!15! 8x! x2

50.x+ 5

x2 + 2x+ 1 ÷x! 2

x2 + 10x+ 9

51.x2 ! 4x+ 8

x2 ! 10x+ 16x+ 3

52.27 ! 6x! x2

x2 + 9x+ 20 ÷x2 ! 12x+ 27x2 + 5x

53.x2 + 5x+ 6x2 ! 36x! 7!6! x

54.2! xx! 5 ÷

x2 + 3x! 10x2 ! 14x+ 48


Version: Fall 2007

55.x+ 3

x2 + 4x! 12x! 4x2 ! 36

56.x+ 3x2 ! x! 2 ÷

x

x2 ! 3x! 4

57.x2 ! 11x+ 28x2 + 5x+ 6 ÷

7x2 ! 30x+ 8x2 ! x! 6

58.x! 73! x

2x2 + 3x! 5x2 ! 12x+ 27

59. Let

f(x) = x2 ! 7x+ 10x2 + 4x! 21

and

g(x) = 5x! x2

x2 + 15x+ 56Compute f(x)/g(x) and simplify your an-swer.

60. Let

f(x) = x2 + 15x+ 56x2 ! x! 20

and

g(x) = !7 ! xx+ 1

Compute f(x)/g(x) and simplify your an-swer.

61. Let

f(x) = x2 + 12x+ 35x2 + 4x! 32

and

g(x) = x2 ! 2x! 35x2 + 8x


62. Let

f(x) = x2 + 4x+ 3x! 1

and

g(x) = x2 ! 4x! 21x+ 5


63. Let

f(x) = x2 + x! 20x

and

g(x) = x! 1x2 ! 2x! 35

Compute f(x)g(x) and simplify your an-swer.

64. Let

f(x) = x2 + 10x+ 24x2 ! 13x+ 42

and

g(x) = x2 ! 6x! 7x2 + 8x+ 12



Version: Fall 2007

65. Let

f(x) = x+ 5!6! x

and

g(x) = x2 + 8x+ 12x2 ! 49


66. Let

f(x) = 8! 7x! x2

x2 ! 8x! 9and

g(x) = x2 ! 6x! 7x2 ! 6x+ 5



Version: Fall 2007

7.4 Solutions

1.10814 ·

6100 = 2 · 2 · 3 · 3 · 3

2 · 7 · 2 · 32 · 2 · 5 · 5 = 3 · 3 · 3 · 3

7 · 5 · 5 = 81175

3.18956 ·

1227 = 3 · 3 · 3 · 7

2 · 2 · 2 · 7 ·2 · 2 · 33 · 3 · 3 = 3

2

5.1536 ·

28100 = 3 · 5

2 · 2 · 3 · 3 ·2 · 2 · 7

2 · 2 · 5 · 5 = 73 · 2 · 2 · 5 = 7

60

7.21100 ·

12516 = 3 · 7

2 · 2 · 5 · 5 ·5 · 5 · 5

2 · 2 · 2 · 2 = 3 · 7 · 52 · 2 · 2 · 2 · 2 · 2 = 105

64

9.5620 ·

9832 = 2 · 2 · 2 · 7

2 · 2 · 5 ·2 · 7 · 7

2 · 2 · 2 · 2 · 2 = 7 · 7 · 75 · 2 · 2 · 2 = 343

40

11. First factor the numerators and the denominators:x+ 6

(x+ 9)(x+ 7) ·x(x+ 7)x+ 4

Then cancel all common factors and multiply the two remaining fractions:x(x+ 6)

(x+ 9)(x+ 4)

Restricted values are !9, !7, and !4.

13. First multiply !9 + 10x! x2 by !1 and also negate the fraction:

!x2 + 7x+ 10x2 ! 1 · x

2 ! 10x+ 9x2 + 9x+ 20

Then factor the numerators and the denominators:

!(x+ 2)(x+ 5)(x! 1)(x+ 1) ·

(x! 1)(x! 9)(x+ 4)(x+ 5)

Section 7.4 Products and Quotients of Rational Functions

Version: Fall 2007

Finally, cancel all common factors and multiply the two remaining fractions:

!(x+ 2)(x! 9)(x+ 1)(x+ 4)

Restricted values are 1, !1, !4, and !5.

15. First factor the numerators and the denominators:x(x! 5)

(x+ 8)(x! 6) ·(x+ 8)(x+ 3)x(x! 1)

Then cancel all common factors and multiply the two remaining fractions:(x! 5)(x+ 3)(x! 6)(x! 1)

Restricted values are !8, 6, 1, and 0.

17. First multiply !x! x2 by !1 and also negate the fraction:

! x2 + xx2 ! 9x+ 8 ·

x2 ! 4x+ 3x2 + 4x+ 3


! x(x+ 1)(x! 1)(x! 8) ·

(x! 1)(x! 3)(x+ 3)(x+ 1)


! x(x! 3)(x! 8)(x+ 3)

Restricted values are 1, 8, !3, and !1.

19. First multiply 7 ! x by !1 and also negate the fraction:

!x+ 2x! 7 ·

x2 + x! 56x2 + 7x+ 6


!x+ 2x! 7 ·

(x! 7)(x+ 8)(x+ 1)(x+ 6)


!(x+ 2)(x+ 8)(x+ 1)(x+ 6)

Restricted values are 7, !1, and !6.


Version: Fall 2007

21. First multiply !3! x by !1 and also negate the fraction:

! x2 ! 9x2 ! 4x! 45 ·

x! 6x+ 3


!(x+ 3)(x! 3)(x+ 5)(x! 9) ·

x! 6x+ 3


!(x! 3)(x! 6)(x+ 5)(x! 9)

Restricted values are !3, !5, and 9.

23. First factor the numerators and the denominators:x+ 5

(x+ 8)(x+ 4) ·(x! 6)(x+ 4)x+ 7

Then cancel all common factors and multiply the two remaining fractions:(x+ 5)(x! 6)(x+ 8)(x+ 7)


25. First factor the numerators and the denominators:x! 5

(x! 2)(x! 6) ·(x! 6)2

x! 8

Then cancel all common factors and multiply the two remaining fractions:(x! 5)(x! 6)(x! 2)(x! 8)

Restricted values are 2, 6, and 8.

27. First factor the numerators and the denominators:(x! 3)(x+ 5)(x! 2)(x! 8) ·

(x! 2)(x! 5)(3x! 2)(x+ 5)

Then cancel all common factors and multiply the two remaining fractions:(x! 3)(x! 5)(3x! 2)(x! 8)

Restricted values are 2, 8, 2/3, and !5.


Version: Fall 2007

29. First factor the numerators and the denominators:(x! 2)(x+ 2)(x+ 9)(x! 7) ·

(x+ 9)(x! 3)(x! 8)(x+ 2)

Then cancel all common factors and multiply the two remaining fractions:(x! 2)(x! 3)(x! 7)(x! 8)

Restricted values are !9, 7, 8, and !2.

31. First factor the numerators and the denominators:x! 1

(x! 7)(x+ 9) ·(x! 9)(x+ 9)x+ 4

Then cancel all common factors and multiply the two remaining fractions:(x! 1)(x! 9)(x! 7)(x+ 4)

Restricted values are 7, !9, and !4.

33. First multiply 5! x by !1 and also negate the fraction:

!x! 5x+ 3 ·

x2 + 3x! 182x2 ! 7x! 15


!x! 5x+ 3 ·

(x+ 6)(x! 3)(2x+ 3)(x! 5)


! (x+ 6)(x! 3)(2x+ 3)(x+ 3)

Restricted values are !3, !3/2, and 5.

35. First rewrite as a multiplication problem:x2 ! 14x+ 48x2 + 10x+ 16 ·

x2 ! x! 72!24 + 11x! x2

Then multiply !24 + 11x! x2 by !1 and also negate the fraction:

!x2 ! 14x+ 48x2 + 10x+ 16 ·

x2 ! x! 72x2 ! 11x+ 24


!(x! 6)(x! 8)(x+ 8)(x+ 2) ·

(x+ 8)(x! 9)(x! 3)(x! 8)


Version: Fall 2007


!(x! 6)(x! 9)(x+ 2)(x! 3)

Restricted values are !8, !2, 9, 3, and 8.

37. First rewrite as a multiplication problem:x2 ! 1

x2 ! 7x+ 12 ·!24 + 10x! x2

x2 + 6x+ 5

Then multiply !24 + 10x! x2 by !1 and also negate the fraction:

! x2 ! 1x2 ! 7x+ 12 ·

x2 ! 10x+ 24x2 + 6x+ 5


!(x! 1)(x+ 1)(x! 4)(x! 3) ·

(x! 4)(x! 6)(x+ 5)(x+ 1)


!(x! 1)(x! 6)(x! 3)(x+ 5)

Restricted values are 4, 3, 6, !5, and !1.

39. First factor the numerators and the denominators, and rewrite as a multiplicationproblem:

(x! 5)(x+ 5)x+ 1 · x! 3

(5x! 2)(x+ 5)

Then cancel all common factors and multiply the two remaining fractions:(x! 5)(x! 3)(5x! 2)(x+ 1)

Restricted values are !1, 2/5, !5, and 3.


(x+ 7)(x+ 3)x! 4 · x+ 8

x(x+ 3)

Then cancel all common factors and multiply the two remaining fractions:(x+ 7)(x+ 8)x(x! 4)



Version: Fall 2007


(x+ 2)(x! 8)(x+ 7)(x! 6) ·

(x+ 7)(x+ 5)(x+ 8)(x! 8)

Then cancel all common factors and multiply the two remaining fractions:(x+ 2)(x+ 5)(x! 6)(x+ 8)

Restricted values are !7, 6, !5, !8, and 8.


(x+ 7)(x+ 5)x+ 4 · x+ 9

(x+ 5)2

Then cancel all common factors and multiply the two remaining fractions:(x+ 7)(x+ 9)(x+ 4)(x+ 5)


47. First rewrite as a multiplication problem:x2 + x! 30x2 + 5x! 36 ·

x+ 8!6! x

Then multiply !6! x by !1 and also negate the fraction:

! x2 + x! 30x2 + 5x! 36 ·

x+ 8x+ 6


!(x+ 6)(x! 5)(x! 4)(x+ 9) ·

x+ 8x+ 6


!(x! 5)(x+ 8)(x! 4)(x+ 9)

Restricted values are 4, !9, !8, and !6.


Version: Fall 2007

49. First rewrite as a multiplication problem:x2 ! 9x+ 8x2 ! 9 · !15! 8x! x2

x2 ! 8xThen multiply !15! 8x! x2 by !1 and also negate the fraction:

!x2 ! 9x+ 8x2 ! 9 · x

2 + 8x+ 15x2 ! 8x


!(x! 1)(x! 8)(x+ 3)(x! 3) ·

(x+ 3)(x+ 5)x(x! 8)


!(x! 1)(x+ 5)x(x! 3)

Restricted values are !3, 3, !5, 0, and 8.


(x+ 2)(x! 2)x+ 8 · x+ 3

(x! 8)(x! 2)Then cancel all common factors and multiply the two remaining fractions:

(x+ 2)(x+ 3)(x+ 8)(x! 8)

Restricted values are !8, 8, 2, and !3.

53. First rewrite as a multiplication problem:x2 + 5x+ 6x2 ! 36 · !6! x

x! 7Then multiply !6! x by !1 and also negate the fraction:

!x2 + 5x+ 6x2 ! 36 · x+ 6

x! 7Then factor the numerators and the denominators:

!(x+ 2)(x+ 3)(x! 6)(x+ 6) ·

x+ 6x! 7


!(x+ 2)(x+ 3)(x! 6)(x! 7)

Restricted values are 6, !6, and 7.


Version: Fall 2007


x+ 3(x! 2)(x+ 6) ·

(x! 6)(x+ 6)x! 4

Then cancel all common factors and multiply the two remaining fractions:(x+ 3)(x! 6)(x! 2)(x! 4)

Restricted values are 2, !6, 4, and 6.


(x! 7)(x! 4)(x+ 2)(x+ 3) ·

(x+ 2)(x! 3)(7x! 2)(x! 4)

Then cancel all common factors and multiply the two remaining fractions:(x! 7)(x! 3)(7x! 2)(x+ 3)

Restricted values are !2, !3, 3, 2/7, and 4.

59.f(x)g(x) = x

2 ! 7x+ 10x2 + 4x! 21 ·

x2 + 15x+ 565x! x2

First multiply 5x! x2 by !1 and also negate the fraction:

!x2 ! 7x+ 10x2 + 4x! 21 ·

x2 + 15x+ 56x2 ! 5x


!(x! 2)(x! 5)(x+ 7)(x! 3) ·

(x+ 7)(x+ 8)x(x! 5)


!(x! 2)(x+ 8)x(x! 3)

Restricted values are !7, 3, !8, 0, and 5.


Version: Fall 2007

61.f(x)g(x) = x

2 + 12x+ 35x2 + 4x! 32 ·

x2 + 8xx2 ! 2x! 35

Factor the numerators and the denominators:(x+ 7)(x+ 5)(x+ 8)(x! 4) ·

x(x+ 8)(x! 7)(x+ 5)

Then cancel all common factors and multiply the two remaining fractions:x(x+ 7)

(x! 4)(x! 7)

Restricted values are !8, 4, 0, 7, and !5.

63.

f(x)g(x) = x2 + x! 20x

· x! 1x2 ! 2x! 35

First factor the numerators and the denominators:(x! 4)(x+ 5)

x· x! 1

(x! 7)(x+ 5)

Then cancel all common factors and multiply the two remaining fractions:(x! 4)(x! 1)x(x! 7)

Restricted values are 0, 7, and !5.

65.

f(x)g(x) = x+ 5!6! x ·

x2 + 8x+ 12x2 ! 49

First multiply !6! x by !1 and also negate the fraction:

!x+ 5x+ 6 ·

x2 + 8x+ 12x2 ! 49


!x+ 5x+ 6 ·

(x+ 6)(x+ 2)(x+ 7)(x! 7)


!(x+ 5)(x+ 2)(x+ 7)(x! 7)


Section 7.5 Sums and Differences of Rational Functions 691

Version: Fall 2007

7.5 Exercises

In Exercises 1-16, add or subtract therational expressions, as indicated, andsimplify your answer. State all restric-tions.

1. 7x2 ! 49xx! 6 + 42

x! 6

2. 2x2 ! 110x! 7 ! 12

7 ! x

3. 27x! 9x2

x+ 3 + 162x+ 3

4. 2x2 ! 28x+ 2 ! 10x

x+ 2

5. 4x2 ! 8x! 4 + 56

4! x

6. 4x2

x! 2 !36x! 56x! 2

7. 9x2

x! 1 + 72x! 631! x

8. 5x2 + 30x! 6 ! 35x

x! 6

9. 4x2 ! 60xx! 7 + 224

x! 7

10. 3x2

x! 7 !63! 30x

7 ! x

11. 3x2

x! 2 !48! 30x

2! x

12. 4x2 ! 164x! 6 ! 20

6! x

13. 9x2

x! 2 !81x! 126x! 2


14. 9x2

x! 8 + 144x! 5768! x

15. 3x2 ! 12x! 3 + 15

3! x

16. 7x2

x! 9 !112x! 441x! 9

In Exercises 17-34, add or subtract therational expressions, as indicated, andsimplify your answer. State all restric-tions.

17. 3xx2 ! 6x+ 5 + 15

x2 ! 14x+ 45

18. 7xx2 ! 4x + 28

x2 ! 12x+ 32

19. 9xx2 + 4x! 12 !

54x2 + 20x+ 84

20. 9xx2 ! 25 !

45x2 + 20x+ 75

21. 5xx2 ! 21x+ 98 !

357x! x2

22. 7x7x! x2 + 147

x2 + 7x! 98

23. !7xx2 ! 8x+ 15 !

35x2 ! 12x+ 35

24. !6xx2 + 2x + 12

x2 + 6x+ 8

25. !9xx2 ! 12x+ 32 !

36x2 ! 4x

26. 5xx2 ! 12x+ 32 !

204x! x2


Version: Fall 2007

27. 6xx2 ! 21x+ 98 !

427x! x2

28. !2xx2 ! 3x! 10 + 4

x2 + 11x+ 18

29. !9xx2 ! 6x+ 8 !

18x2 ! 2x

30. 6x5x! x2 + 90

x2 + 5x! 50

31. 8x5x! x2 + 120

x2 + 5x! 50

32. !5xx2 + 5x + 25

x2 + 15x+ 50

33. !5xx2 + x! 30 + 30

x2 + 23x+ 102

34. 9xx2 + 12x+ 32 !

36x2 + 4x

35. Let

f(x) = 8xx2 + 6x+ 8

and

g(x) = 16x2 + 2x

Compute f(x) ! g(x) and simplify youranswer.

36. Let

f(x) = !7xx2 + 8x+ 12

and

g(x) = 42x2 + 16x+ 60

Compute f(x) + g(x) and simplify youranswer.

37. Let

f(x) = 11xx2 + 12x+ 32

and

g(x) = 44!4x! x2


38. Let

f(x) = 8xx2 ! 6x

and

g(x) = 48x2 ! 18x+ 72


39. Let

f(x) = 4x!x! x2

and

g(x) = 4x2 + 3x+ 2


40. Let

f(x) = 5xx2 ! x! 12

and

g(x) = 15x2 + 13x+ 30

Compute f(x) ! g(x) and simplify youranswer.

Section 7.5 Sums and Differences of Rational Functions

Version: Fall 2007

7.5 Solutions

1. Provided x "= 6,7x2 ! 49xx! 6 + 42

x! 6 = 7x2 ! 49x+ 42x! 6

= 7(x2 ! 7x+ 6)x! 6

= 7(x! 6)(x! 1)x! 6

= 7(x! 1)

3. Provided x "= !3,27x! 9x2

x+ 3 + 162x+ 3 = !9x2 + 27x+ 162

x+ 3

= !9(x2 ! 3x! 18)x+ 3

= !9(x+ 3)(x! 6)x+ 3

= !9(x! 6)

5. Provided x "= 4,4x2 ! 8x! 4 + 56

4! x = 4x2 ! 8x! 4 !

56x! 4

= 4x2 ! 8! 56x! 4

= 4x2 ! 64x! 4

= 4(x2 ! 16)x! 4

= 4(x! 4)(x+ 4)x! 4

= 4(x+ 4)


Version: Fall 2007

7. Provided x "= 1,9x2

x! 1 + 72x! 631! x = 9x2

x! 1 !72x! 63x! 1

= 9x2 ! 72x+ 63x! 1

= 9(x2 ! 8x+ 7)x! 1

= 9(x! 1)(x! 7)x! 1

= 9(x! 7)

9. Provided x "= 7,4x2 ! 60xx! 7 + 224

x! 7 = 4x2 ! 60x+ 224x! 7

= 4(x2 ! 15x+ 56)x! 7

= 4(x! 7)(x! 8)x! 7

= 4(x! 8)


x! 2 !48! 30x

2! x = 3x2

x! 2 + 48! 30xx! 2

= 3x2 ! 30x+ 48x! 2

= 3(x2 ! 10x+ 16)x! 2

= 3(x! 2)(x! 8)x! 2

= 3(x! 8)


x! 2 !81x! 126x! 2 = 9x2 ! 81x+ 126

x! 2

= 9(x2 ! 9x+ 14)x! 2

= 9(x! 2)(x! 7)x! 2

= 9(x! 7)


Version: Fall 2007

15. Provided x "= 3,3x2 ! 12x! 3 + 15

3! x = 3x2 ! 12x! 3 ! 15

x! 3

= 3x2 ! 12! 15x! 3

= 3x2 ! 27x! 3

= 3(x2 ! 9)x! 3

= 3(x! 3)(x+ 3)x! 3

= 3(x+ 3)

17.3x

x2 ! 6x+ 5 + 15x2 ! 14x+ 45

= 3x(x! 5)(x! 1) + 15

(x! 5)(x! 9)

= 3x(x! 9)(x! 5)(x! 1)(x! 9) + 15(x! 1)

(x! 5)(x! 1)(x! 9)

= 3x(x! 9) + 15(x! 1)(x! 5)(x! 1)(x! 9)

= 3x2 ! 12x! 15(x! 5)(x! 1)(x! 9)

= 3(x! 5)(x+ 1)(x! 5)(x! 1)(x! 9)

= 3(x+ 1)(x! 1)(x! 9)



Version: Fall 2007

19.9x

x2 + 4x! 12 !54

x2 + 20x+ 84

= 9x(x+ 6)(x! 2) !

54(x+ 6)(x+ 14)

= 9x(x+ 14)(x+ 6)(x! 2)(x+ 14) !

54(x! 2)(x+ 6)(x! 2)(x+ 14)

= 9x(x+ 14)! 54(x! 2)(x+ 6)(x! 2)(x+ 14)

= 9x2 + 72x+ 108(x+ 6)(x! 2)(x+ 14)

= 9(x+ 6)(x+ 2)(x+ 6)(x! 2)(x+ 14)

= 9(x+ 2)(x! 2)(x+ 14)

Restricted values are !6, 2, and !14.

21.5x

x2 ! 21x+ 98 !35

7x! x2

= 5xx2 ! 21x+ 98 + 35

x2 ! 7x

= 5x(x! 7)(x! 14) + 35

x(x! 7)

= 5x2

x(x! 7)(x! 14) + 35(x! 14)x(x! 7)(x! 14)

= 5x2 + 35(x! 14)x(x! 7)(x! 14)

= 5x2 + 35x! 490x(x! 7)(x! 14)

= 5(x! 7)(x+ 14)x(x! 7)(x! 14)

= 5(x+ 14)x(x! 14)



Version: Fall 2007

23.!7x

x2 ! 8x+ 15 !35

x2 ! 12x+ 35

= !7x(x! 5)(x! 3) !

35(x! 5)(x! 7)

= !7x(x! 7)(x! 5)(x! 3)(x! 7) !

35(x! 3)(x! 5)(x! 3)(x! 7)

= !7x(x! 7)! 35(x! 3)(x! 5)(x! 3)(x! 7)

= !7x2 + 14x+ 105(x! 5)(x! 3)(x! 7)

= !7(x! 5)(x+ 3)(x! 5)(x! 3)(x! 7)

= !7(x+ 3)(x! 3)(x! 7)


25.!9x

x2 ! 12x+ 32 !36

x2 ! 4x

= !9x(x! 4)(x! 8) !

36x(x! 4)

= !9x2

x(x! 4)(x! 8) !36(x! 8)

x(x! 4)(x! 8)

= !9x2 ! 36(x! 8)x(x! 4)(x! 8)

= !9x2 ! 36x+ 288x(x! 4)(x! 8)

= !9(x! 4)(x+ 8)x(x! 4)(x! 8)

= !9(x+ 8)x(x! 8)



Version: Fall 2007

27.6x

x2 ! 21x+ 98 !42

7x! x2

= 6xx2 ! 21x+ 98 + 42

x2 ! 7x

= 6x(x! 7)(x! 14) + 42

x(x! 7)

= 6x2

x(x! 7)(x! 14) + 42(x! 14)x(x! 7)(x! 14)

= 6x2 + 42(x! 14)x(x! 7)(x! 14)

= 6x2 + 42x! 588x(x! 7)(x! 14)

= 6(x! 7)(x+ 14)x(x! 7)(x! 14)

= 6(x+ 14)x(x! 14)


29.!9x

x2 ! 6x+ 8 !18

x2 ! 2x

= !9x(x! 2)(x! 4) !

18x(x! 2)

= !9x2

x(x! 2)(x! 4) !18(x! 4)

x(x! 2)(x! 4)

= !9x2 ! 18(x! 4)x(x! 2)(x! 4)

= !9x2 ! 18x+ 72x(x! 2)(x! 4)

= !9(x! 2)(x+ 4)x(x! 2)(x! 4)

= !9(x+ 4)x(x! 4)



Version: Fall 2007

31.8x

5x! x2 + 120x2 + 5x! 50

= !8xx2 ! 5x + 120

x2 + 5x! 50

= !8xx(x! 5) + 120

(x! 5)(x+ 10)

= !8x(x+ 10)x(x! 5)(x+ 10) + 120x

x(x! 5)(x+ 10)

= !8x(x+ 10) + 120xx(x! 5)(x+ 10)

= !8x2 + 40xx(x! 5)(x+ 10)

= !8x(x! 5)x(x! 5)(x+ 10)

= !8x+ 10

Restricted values are 5, 0, and !10.

33.!5x

x2 + x! 30 + 30x2 + 23x+ 102

= !5x(x+ 6)(x! 5) + 30

(x+ 6)(x+ 17)

= !5x(x+ 17)(x+ 6)(x! 5)(x+ 17) + 30(x! 5)

(x+ 6)(x! 5)(x+ 17)

= !5x(x+ 17) + 30(x! 5)(x+ 6)(x! 5)(x+ 17)

= !5x2 ! 55x! 150(x+ 6)(x! 5)(x+ 17)

= !5(x+ 6)(x+ 5)(x+ 6)(x! 5)(x+ 17)

= !5(x+ 5)(x! 5)(x+ 17)



Version: Fall 2007

35.

f(x)! g(x) = 8xx2 + 6x+ 8 !

16x2 + 2x

= 8x(x+ 2)(x+ 4) !

16x(x+ 2)

= 8x2

x(x+ 2)(x+ 4) !16(x+ 4)

x(x+ 2)(x+ 4)

= 8x2 ! 16(x+ 4)x(x+ 2)(x+ 4)

= 8x2 ! 16x! 64x(x+ 2)(x+ 4)

= 8(x+ 2)(x! 4)x(x+ 2)(x+ 4)

= 8(x! 4)x(x+ 4)


37.

f(x) + g(x) = 11xx2 + 12x+ 32 + 44

!4x! x2

= 11xx2 + 12x+ 32 !

44x2 + 4x

= 11x(x+ 4)(x+ 8) !

44x(x+ 4)

= 11x2

x(x+ 4)(x+ 8) !44(x+ 8)

x(x+ 4)(x+ 8)

= 11x2 ! 44(x+ 8)x(x+ 4)(x+ 8)

= 11x2 ! 44x! 352x(x+ 4)(x+ 8)

= 11(x+ 4)(x! 8)x(x+ 4)(x+ 8)

= 11(x! 8)x(x+ 8)



Version: Fall 2007

39.

f(x) + g(x) = 4x!x! x2 + 4

x2 + 3x+ 2

= !4xx2 + x + 4

x2 + 3x+ 2

= !4xx(x+ 1) + 4

(x+ 1)(x+ 2)

= !4x(x+ 2)x(x+ 1)(x+ 2) + 4x

x(x+ 1)(x+ 2)

= !4x(x+ 2) + 4xx(x+ 1)(x+ 2)

= !4x2 ! 4xx(x+ 1)(x+ 2)

= !4x(x+ 1)x(x+ 1)(x+ 2)

= !4xx(x+ 2)

= !4x+ 2


Section 7.6 Complex Fractions 703

Version: Fall 2007

7.6 Exercises

In Exercises 1-6, evaluate the functionat the given rational number. Then usethe first or second technique for simpli-fying complex fractions explained in thenarrative to simplify your answer.

1. Given

f(x) = x+ 12! x,

evaluate and simplify f(1/2).

2. Given

f(x) = 2! xx+ 5 ,


3. Given

f(x) = 2x+ 34! x ,


4. Given

f(x) = 3! 2xx+ 5 ,


5. Given

f(x) = 5! 2xx+ 4 ,


6. Given

f(x) = 2x! 911! x,



In Exercises 7-46, simplify the givencomplex rational expression. State all re-strictions.

7.

5 + 6x

25x! 36x3

8.

7 + 9x

49x! 81x3

9.7x! 2 !

5x! 7

8x! 7 + 3

x+ 8

10.9x+ 4 !

7x! 9

9x! 9 + 5

x! 4

11.

3 + 7x

9x2 !

49x4


Version: Fall 2007

12.

2! 5x

4x2 !

25x4

13.9x+ 4 + 7

x+ 99x+ 9 + 2

x! 8

14.4x! 6 + 9

x! 99x! 6 + 8

x! 9

15.5x! 7 !

4x! 4

10x! 4 !

5x+ 2

16.3x+ 6 + 7

x+ 99x+ 6 !

4x+ 9

17.6x! 3 + 5

x! 89x! 3 + 7

x! 8

18.7x! 7 !

4x! 2

7x! 7 !

6x! 2

19.4x! 2 + 7

x! 75x! 2 + 2

x! 7

20.9x+ 2 !

7x+ 5

4x+ 2 + 3

x+ 5

21.

5 + 4x

25x! 16x3

22.6x+ 5 + 5

x+ 48x+ 5 !

3x+ 4

23.9x! 5 + 8

x+ 45x! 5 !

4x+ 4

24.4x! 6 + 4

x! 96x! 6 + 6

x! 9

25.6x+ 8 + 5

x! 25x! 2 !

2x+ 2


Version: Fall 2007

26.7x+ 9 + 9

x! 24x! 2 + 7

x+ 1

27.7x+ 7 !

5x+ 4

8x+ 7 !

3x+ 4

28.

25! 16x2

5 + 4x

29.64x! 25x3

8! 5x

30.4x+ 2 + 5

x! 67x! 6 !

5x+ 7

31.2x! 6 !

4x+ 9

3x! 6 !

6x+ 9

32.3x+ 6 !

4x+ 4

6x+ 6 !

8x+ 4

33.9x2 !

64x4

3! 8x

34.9x2 !

25x4

3! 5x

35.4x! 4 !

8x! 7

4x! 7 + 2

x+ 2

36.

2! 7x

4! 49x2

37.3

x2 + 8x! 9 + 3x2 ! 81

9x2 ! 81 + 9

x2 ! 8x! 9

38.7

x2 ! 5x! 14 + 2x2 ! 7x! 18

5x2 ! 7x! 18 + 8

x2 ! 6x! 27

39.2

x2 + 8x+ 7 + 5x2 + 13x+ 42

7x2 + 13x+ 42 + 6

x2 + 3x! 18


Version: Fall 2007

40.3

x2 + 5x! 14 + 3x2 ! 7x! 98

3x2 ! 7x! 98 + 3

x2 ! 15x+ 14

41.6

x2 + 11x+ 24 !6

x2 + 13x+ 409

x2 + 13x+ 40 !9

x2 ! 3x! 40

42.7

x2 + 13x+ 30 + 7x2 + 19x+ 90

9x2 + 19x+ 90 + 9

x2 + 7x! 18

43.7

x2 ! 6x+ 5 + 7x2 + 2x! 35

8x2 + 2x! 35 + 8

x2 + 8x+ 7

44.2

x2 ! 4x! 12 !2

x2 ! x! 302

x2 ! x! 30 !2

x2 ! 4x! 45

45.4

x2 + 6x! 7 !4

x2 + 2x! 34

x2 + 2x! 3 !4

x2 + 5x+ 6

46.9

x2 + 3x! 4 + 8x2 ! 7x+ 6

4x2 ! 7x+ 6 + 9

x2 ! 10x+ 24

47. Given f(x) = 2/x, simplifyf(x)! f(3)x! 3 .

State all restrictions.

48. Given f(x) = 5/x, simplifyf(x)! f(2)x! 2 .


49. Given f(x) = 3/x2, simplifyf(x)! f(1)x! 1 .


50. Given f(x) = 5/x2, simplifyf(x)! f(2)x! 2 .


51. Given f(x) = 7/x, simplifyf(x+ h)! f(x)

h.


52. Given f(x) = 4/x, simplifyf(x+ h)! f(x)

h.


53. Given

f(x) = x+ 13! x,

find and simplify f(1/x). State all re-strictions.


Version: Fall 2007

54. Given

f(x) = 2! x3x+ 4 ,


55. Given

f(x) = x+ 12! 5x,


56. Given

f(x) = 2x! 34 + x ,


57. Given

f(x) = x

x+ 2 ,

find and simplify f(f(x)). State all re-strictions.

58. Given

f(x) = 2xx+ 5 ,

find and simplify f(f(x)). State all re-strictions.


Version: Fall 2007

7.6 Solutions

1.

f(1/2) =12 + 12! 1

2=

12 + 12! 1

2· 22 = 1 + 2

4! 1 = 33 = 1

3.

f(1/3) =2!1

3"

+ 34! 1

3=

2!1

3"

+ 34! 1

3· 33 = 2 + 9

12! 1 = 1111 = 1

5.

f(3/5) =5! 2

!35"

35 + 4

=5! 2

!35"

35 + 4

· 55 = 25! 63 + 20 = 19

23

7. First multiply the main numerator and denominator by the least common denom-inator (LCD) of the two small fractions, which is x3:

#5 + 6x

$· x

3

1#25x! 36x3

$· x

3

1

= 5x3 + 6x2

25x2 ! 36

Then factor the numerator and denominator:x2(5x+ 6)

(5x+ 6)(5x! 6)Finally, cancel all common factors in the numerator and denominator:

x2

5x! 6Restricted values are 0, !6/5, and 6/5.

9. First multiply the main numerator and denominator by the least common denom-inator (LCD) of the two small fractions, which is (x! 2)(x! 7)(x+ 8):

# 7x! 2 !

5x! 7

$· (x! 2)(x! 7)(x+ 8)

1# 8x! 7 + 3

x+ 8

$· (x! 2)(x! 7)(x+ 8)

1

= 7(x! 7)(x+ 8)! 5(x! 2)(x+ 8)8(x! 2)(x+ 8) + 3(x! 2)(x! 7)

Section 7.6 Complex Fractions

Version: Fall 2007

Then simplify the numerator and denominator:7(x! 7)(x+ 8)! 5(x! 2)(x+ 8)8(x! 2)(x+ 8) + 3(x! 2)(x! 7) = (7(x! 7)! 5(x! 2))(x+ 8)

(8(x+ 8) + 3(x! 7))(x! 2)

= (2x! 39)(x+ 8)(11x+ 43)(x! 2)

Finally, cancel all common factors, if any, in the numerator and denominator. In thiscase, there are no common factors.Restricted values are 2, 7, !8, and !43/11.

11. First multiply the main numerator and denominator by the least common de-nominator (LCD) of the two small fractions, which is x4:

#3 + 7x

$· x

4

1# 9x2 !

49x4

$· x

4

1

= 3x4 + 7x3

9x2 ! 49


(3x+ 7)(3x! 7)

Finally, cancel all common factors in the numerator and denominator:x3


13. First multiply the main numerator and denominator by the least common de-nominator (LCD) of the two small fractions, which is (x+ 4)(x+ 9)(x! 8):

# 9x+ 4 + 7

x+ 9

$· (x+ 4)(x+ 9)(x! 8)

1# 9x+ 9 + 2

x! 8

$· (x+ 4)(x+ 9)(x! 8)

1

= 9(x+ 9)(x! 8) + 7(x+ 4)(x! 8)9(x+ 4)(x! 8) + 2(x+ 4)(x+ 9)

Then simplify the numerator and denominator:9(x+ 9)(x! 8) + 7(x+ 4)(x! 8)9(x+ 4)(x! 8) + 2(x+ 4)(x+ 9) = (9(x+ 9) + 7(x+ 4))(x! 8)

(9(x! 8) + 2(x+ 9))(x+ 4)

= (16x+ 109)(x! 8)(11x! 54)(x+ 4)


Version: Fall 2007

Finally, cancel all common factors, if any, in the numerator and denominator. In thiscase, there are no common factors.Restricted values are !4, !9, 8, and 54/11.

15. First multiply the main numerator and denominator by the least common de-nominator (LCD) of the two small fractions, which is (x! 7)(x! 4)(x+ 2):

# 5x! 7 !

4x! 4

$· (x! 7)(x! 4)(x+ 2)

1# 10x! 4 !

5x+ 2

$· (x! 7)(x! 4)(x+ 2)

1

= 5(x! 4)(x+ 2)! 4(x! 7)(x+ 2)10(x! 7)(x+ 2)! 5(x! 7)(x! 4)

Then simplify the numerator and denominator:5(x! 4)(x+ 2)! 4(x! 7)(x+ 2)10(x! 7)(x+ 2)! 5(x! 7)(x! 4) = (5(x! 4)! 4(x! 7))(x+ 2)

(10(x+ 2)! 5(x! 4))(x! 7)

= (x+ 8)(x+ 2)(5x+ 40)(x! 7)

Finally, cancel all common factors, if any, in the numerator and denominator:(x+ 8)(x+ 2)

(5x+ 40)(x! 7) = x+ 25(x! 7)


17. First multiply the main numerator and denominator by the least common de-nominator (LCD) of the two small fractions, which is (x! 3)(x! 8):

# 6x! 3 + 5

x! 8

$· (x! 3)(x! 8)

1# 9x! 3 + 7

x! 8

$· (x! 3)(x! 8)

1

= 6(x! 8) + 5(x! 3)9(x! 8) + 7(x! 3)

Then simplify the numerator and denominator:6(x! 8) + 5(x! 3)9(x! 8) + 7(x! 3) = 11x! 63

16x! 93

Finally, cancel all common factors, if any, in the numerator and denominator. In thiscase, there are no common factors.Restricted values are 3, 8, and 93/16.


Version: Fall 2007

19. First multiply the main numerator and denominator by the least common de-nominator (LCD) of the two small fractions, which is (x! 2)(x! 7):

# 4x! 2 + 7

x! 7

$· (x! 2)(x! 7)

1# 5x! 2 + 2

x! 7

$· (x! 2)(x! 7)

1

= 4(x! 7) + 7(x! 2)5(x! 7) + 2(x! 2)

Then simplify the numerator and denominator:4(x! 7) + 7(x! 2)5(x! 7) + 2(x! 2) = 11x! 42

7x! 39

Finally, cancel all common factors, if any, in the numerator and denominator. In thiscase, there are no common factors.Restricted values are 2, 7, and 39/7.


#5 + 4x

$· x

3

1#25x! 16x3

$· x

3

1

= 5x3 + 4x2

25x2 ! 16


(5x+ 4)(5x! 4)

Finally, cancel all common factors in the numerator and denominator:x2


23. First multiply the main numerator and denominator by the least common de-nominator (LCD) of the two small fractions, which is (x! 5)(x+ 4):

# 9x! 5 + 8

x+ 4

$· (x! 5)(x+ 4)

1# 5x! 5 !

4x+ 4

$· (x! 5)(x+ 4)

1

= 9(x+ 4) + 8(x! 5)5(x+ 4)! 4(x! 5)

Then simplify the numerator and denominator:9(x+ 4) + 8(x! 5)5(x+ 4)! 4(x! 5) = 17x! 4

x+ 40


Version: Fall 2007

Finally, cancel all common factors, if any, in the numerator and denominator. In thiscase, there are no common factors.Restricted values are 5, !4, and !40.

25. First multiply the main numerator and denominator by the least common de-nominator (LCD) of the two small fractions, which is (x+ 8)(x! 2)(x+ 2):

# 6x+ 8 + 5

x! 2

$· (x+ 8)(x! 2)(x+ 2)

1# 5x! 2 !

2x+ 2

$· (x+ 8)(x! 2)(x+ 2)

1

= 6(x! 2)(x+ 2) + 5(x+ 8)(x+ 2)5(x+ 8)(x+ 2)! 2(x+ 8)(x! 2)

Then simplify the numerator and denominator:6(x! 2)(x+ 2) + 5(x+ 8)(x+ 2)5(x+ 8)(x+ 2)! 2(x+ 8)(x! 2) = (6(x! 2) + 5(x+ 8))(x+ 2)

(5(x+ 2)! 2(x! 2))(x+ 8)

= (11x+ 28)(x+ 2)(3x+ 14)(x+ 8)

Finally, cancel all common factors, if any, in the numerator and denominator. In thiscase, there are no common factors.Restricted values are !8, 2, !2, and !14/3.

27. First multiply the main numerator and denominator by the least common de-nominator (LCD) of the two small fractions, which is (x+ 7)(x+ 4):

# 7x+ 7 !

5x+ 4

$· (x+ 7)(x+ 4)

1# 8x+ 7 !

3x+ 4

$· (x+ 7)(x+ 4)

1

= 7(x+ 4)! 5(x+ 7)8(x+ 4)! 3(x+ 7)

Then simplify the numerator and denominator:7(x+ 4)! 5(x+ 7)8(x+ 4)! 3(x+ 7) = 2x! 7

5x+ 11

Finally, cancel all common factors, if any, in the numerator and denominator. In thiscase, there are no common factors.Restricted values are !7, !4, and !11/5.


Version: Fall 2007


#64x! 25x3

$· x

3

1#

8! 5x

$· x

3

1

= 64x2 ! 258x3 ! 5x2

Then factor the numerator and denominator:(8x! 5)(8x+ 5)x2(8x! 5)

Finally, cancel all common factors in the numerator and denominator:8x+ 5x2

Restricted values are 0 and 5/8.

31. First multiply the main numerator and denominator by the least common de-nominator (LCD) of the two small fractions, which is (x! 6)(x+ 9):

# 2x! 6 !

4x+ 9

$· (x! 6)(x+ 9)

1# 3x! 6 !

6x+ 9

$· (x! 6)(x+ 9)

1

= 2(x+ 9)! 4(x! 6)3(x+ 9)! 6(x! 6)

Then simplify the numerator and denominator:2(x+ 9)! 4(x! 6)3(x+ 9)! 6(x! 6) = !2x+ 42

!3x+ 63

Finally, cancel all common factors, if any, in the numerator and denominator:!2x+ 42!3x+ 63 = 2

3Restricted values are 6, !9, and 21.


Version: Fall 2007


# 9x2 !

64x4

$· x

4

1#

3! 8x

$· x

4

1

= 9x2 ! 643x4 ! 8x3

Then factor the numerator and denominator:(3x! 8)(3x+ 8)x3(3x! 8)

Finally, cancel all common factors in the numerator and denominator:3x+ 8x3


35. First multiply the main numerator and denominator by the least common de-nominator (LCD) of the two small fractions, which is (x! 4)(x! 7)(x+ 2):

# 4x! 4 !

8x! 7

$· (x! 4)(x! 7)(x+ 2)

1# 4x! 7 + 2

x+ 2

$· (x! 4)(x! 7)(x+ 2)

1

= 4(x! 7)(x+ 2)! 8(x! 4)(x+ 2)4(x! 4)(x+ 2) + 2(x! 4)(x! 7)

Then simplify the numerator and denominator:4(x! 7)(x+ 2)! 8(x! 4)(x+ 2)4(x! 4)(x+ 2) + 2(x! 4)(x! 7) = (4(x! 7)! 8(x! 4))(x+ 2)

(4(x+ 2) + 2(x! 7))(x! 4)

= (!4x+ 4)(x+ 2)(6x! 6)(x! 4)

Finally, cancel all common factors, if any, in the numerator and denominator:(!4x+ 4)(x+ 2)(6x! 6)(x! 4) = !2(x+ 2)

3(x! 4)

Restricted values are 4, 7, !2, and 1.


Version: Fall 2007

37. First factor all of the denominators:3

(x! 1)(x+ 9) + 3(x+ 9)(x! 9)

9(x+ 9)(x! 9) + 9

(x! 9)(x+ 1)

Then simplify the main numerator and denominator:3(x! 9)

(x! 1)(x+ 9)(x! 9) + 3(x! 1)(x! 1)(x+ 9)(x! 9)

9(x+ 1)(x+ 9)(x! 9)(x+ 1) + 9(x+ 9)

(x+ 9)(x! 9)(x+ 1)

=

3(x! 9) + 3(x! 1)(x! 1)(x+ 9)(x! 9)9(x+ 1) + 9(x+ 9)

(x+ 9)(x! 9)(x+ 1)

=

6x! 30(x! 1)(x+ 9)(x! 9)

18x+ 90(x+ 9)(x! 9)(x+ 1)

Then divide by rewriting as a multiplication problem, and cancel common factors:6x! 30

(x! 1)(x+ 9)(x! 9) ·(x+ 9)(x! 9)(x+ 1)

18x+ 90

= (6x! 30)(x+ 1)(18x+ 90)(x! 1)

Finally, factor the numerator and denominator further, if possible, and cancel all com-mon factors again:

6(x! 5)(x+ 1)18(x+ 5)(x! 1)

= (x! 5)(x+ 1)3(x+ 5)(x! 1)

Restricted values are 1, !9, 9, !1, and !5.


Version: Fall 2007


(x+ 1)(x+ 7) + 5(x+ 7)(x+ 6)

7(x+ 7)(x+ 6) + 6

(x+ 6)(x! 3)

Then simplify the main numerator and denominator:2(x+ 6)

(x+ 1)(x+ 7)(x+ 6) + 5(x+ 1)(x+ 1)(x+ 7)(x+ 6)

7(x! 3)(x+ 7)(x+ 6)(x! 3) + 6(x+ 7)

(x+ 7)(x+ 6)(x! 3)

=

2(x+ 6) + 5(x+ 1)(x+ 1)(x+ 7)(x+ 6)7(x! 3) + 6(x+ 7)

(x+ 7)(x+ 6)(x! 3)

=

7x+ 17(x+ 1)(x+ 7)(x+ 6)

13x+ 21(x+ 7)(x+ 6)(x! 3)

Then divide by rewriting as a multiplication problem, and cancel common factors:7x+ 17

(x+ 1)(x+ 7)(x+ 6) ·(x+ 7)(x+ 6)(x! 3)

13x+ 21

= (7x+ 17)(x! 3)(13x+ 21)(x+ 1)

Finally, factor the numerator and denominator further, if possible, and cancel all com-mon factors again, if any. In this case, the expression does not simplify any further.Restricted values are !1, !7, !6, 3, and !21/13.


Version: Fall 2007


(x+ 3)(x+ 8) !6

(x+ 8)(x+ 5)9

(x+ 8)(x+ 5) !9

(x+ 5)(x! 8)


(x+ 3)(x+ 8)(x+ 5) !6(x+ 3)

(x+ 3)(x+ 8)(x+ 5)9(x! 8)

(x+ 8)(x+ 5)(x! 8) !9(x+ 8)

(x+ 8)(x+ 5)(x! 8)

=

6(x+ 5)! 6(x+ 3)(x+ 3)(x+ 8)(x+ 5)9(x! 8)! 9(x+ 8)

(x+ 8)(x+ 5)(x! 8)

=

12(x+ 3)(x+ 8)(x+ 5)

!144(x+ 8)(x+ 5)(x! 8)

Then divide by rewriting as a multiplication problem, and cancel common factors:12

(x+ 3)(x+ 8)(x+ 5) ·(x+ 8)(x+ 5)(x! 8)

!144

= !1(x! 8)12(x+ 3)

Restricted values are !3, !8, !5, and 8.


Version: Fall 2007


(x! 1)(x! 5) + 7(x! 5)(x+ 7)

8(x! 5)(x+ 7) + 8

(x+ 7)(x+ 1)


(x! 1)(x! 5)(x+ 7) + 7(x! 1)(x! 1)(x! 5)(x+ 7)

8(x+ 1)(x! 5)(x+ 7)(x+ 1) + 8(x! 5)

(x! 5)(x+ 7)(x+ 1)

=

7(x+ 7) + 7(x! 1)(x! 1)(x! 5)(x+ 7)8(x+ 1) + 8(x! 5)

(x! 5)(x+ 7)(x+ 1)

=

14x+ 42(x! 1)(x! 5)(x+ 7)

16x! 32(x! 5)(x+ 7)(x+ 1)

Then divide by rewriting as a multiplication problem, and cancel common factors:14x+ 42

(x! 1)(x! 5)(x+ 7) ·(x! 5)(x+ 7)(x+ 1)

16x! 32

= (14x+ 42)(x+ 1)(16x! 32)(x! 1)

Finally, factor the numerator and denominator further, if possible, and cancel all com-mon factors again:

14(x+ 3)(x+ 1)16(x! 2)(x! 1)

= 7(x+ 3)(x+ 1)8(x! 2)(x! 1)

Restricted values are 1, 5, !7, !1, and 2.


Version: Fall 2007


(x+ 7)(x! 1) !4

(x! 1)(x+ 3)4

(x! 1)(x+ 3) !4

(x+ 3)(x+ 2)


(x+ 7)(x! 1)(x+ 3) !4(x+ 7)

(x+ 7)(x! 1)(x+ 3)4(x+ 2)

(x! 1)(x+ 3)(x+ 2) !4(x! 1)

(x! 1)(x+ 3)(x+ 2)

=

4(x+ 3)! 4(x+ 7)(x+ 7)(x! 1)(x+ 3)4(x+ 2)! 4(x! 1)

(x! 1)(x+ 3)(x+ 2)

=

!16(x+ 7)(x! 1)(x+ 3)

12(x! 1)(x+ 3)(x+ 2)

Then divide by rewriting as a multiplication problem, and cancel common factors:!16

(x+ 7)(x! 1)(x+ 3) ·(x! 1)(x+ 3)(x+ 2)

12

= !4(x+ 2)3(x+ 7)

Restricted values are !7, 1, !3, and !2.

47.

f(x)! f(3)x! 3 =

2x !

23

x! 3 =2x !

23

x! 3 ·3x3x = 6! 2x

3x(x! 3) = !2(x! 3)3x(x! 3) = ! 2

3x

Restricted values are 0 and 3.

49.

f(x)! f(1)x! 1 =

3x2 ! 3x! 1 =

3x2 ! 3x! 1 ·

x2

x2 = 3! 3x2

x2(x! 1) = !3(x! 1)(x+ 1)x2(x! 1) = !3(x+ 1)

x2

Restricted values are 0 and 1.


Version: Fall 2007

51.

f(x+ h)! f(x)h

=7x+h !

7x

h=

7x+h !

7x

h·x(x+ h)x(x+ h) = 7x! 7(x+ h)

hx(x+ h) = !7hhx(x+ h) = ! 7

h(x+ h)

Restricted values are x "= 0, x "= !h, and h "= 0.

53.

f(1/x) =1x + 13! 1

x

=1x + 13! 1

x

· xx

= 1 + x3x! 1 = x+ 1

3x! 1


55.

f(5/x) =5x + 1

2! 5! 5x

" =5x + 1

2! 5! 5x

" · xx

= 5 + x2x! 25 = x+ 5

2x! 25


57.

f(f(x)) = f#x

x+ 2

$=

xx+2xx+2 + 2 =

xx+2xx+2 + 2 ·

x+ 2x+ 2 = x

x+ 2(x+ 2) = x

3x+ 4

Restricted values are !2 and !4/3.

Section 7.7 Solving Rational Equations 723

Version: Fall 2007

7.7 Exercises

For each of the rational functions givenin Exercises 1-6, perform each of thefollowing tasks.


ii. Plot the zero of the rational functionon your coordinate system and labelit with its coordinates. Plot the verti-cal and horizontal asymptotes on yourcoordinate system and label them withtheir equations. Use this informa-tion (and your graphing calculator)to draw the graph of f .

iii. Plot the horizontal line y = k on yourcoordinate system and label this linewith its equation.

iv. Use your calculator’s intersect util-ity to help determine the solution off(x) = k. Label this point on yourgraph with its coordinates.

v. Solve the equation f(x) = k alge-braically, placing the work for thissolution on your graph paper next toyour coordinate system containing thegraphical solution. Do the answersagree?

1. f(x) = x! 1x+ 2 ; k = 3

2. f(x) = x+ 1x! 2 ; k = !3

3. f(x) = x+ 13! x ; k = 2

4. f(x) = x+ 32! x ; k = 2

5. f(x) = 2x+ 3x! 1 ; k = !3


6. f(x) = 5! 2xx! 1 ; k = 3

In Exercises 7-14, use a strictly alge-braic technique to solve the equation f(x) =k for the given function and value of k.You are encouraged to check your resultwith your calculator.

7. f(x) = 16x! 92x! 1 ; k = 8

8. f(x) = 10x! 37x+ 7 ; k = 1

9. f(x) = 5x+ 84x+ 1 ; k = !11

10. f(x) = !6x! 117x! 2 ; k = !6

11. f(x) = ! 35x7x+ 12 ; k = !5

12. f(x) = !66x! 56x! 10 ; k = !11

13. f(x) = 8x+ 2x! 11 ; k = 11

14. f(x) = 36x! 73x! 4 ; k = 12

In Exercises 15-20, use a strictly alge-braic technique to solve the given equa-tion. You are encouraged to check yourresult with your calculator.

15. x7 + 89 = !8

7

16. x3 + 92 = !3

8


Version: Fall 2007

17. !57x

= 27 ! 40x2

18. !117x

= 54 + 54x2

19. 7x

= 4! 3x2

20. 3x2 = 5! 3

x

For each of the rational functions givenin Exercises 21-26, perform each of thefollowing tasks.


ii. Plot the zero of the rational functionon your coordinate system and labelit with its coordinates. You may useyour calculator’s zero utility to findthis, if you wish.

iii. Plot the vertical and horizontal as-ymptotes on your coordinate systemand label them with their equations.Use the asymptote and zero informa-tion (and your graphing calculator)to draw the graph of f .

iv. Plot the horizontal line y = k on yourcoordinate system and label this linewith its equation.

v. Use your calculator’s intersect util-ity to help determine the solution off(x) = k. Label this point on yourgraph with its coordinates.

vi. Solve the equation f(x) = k alge-braically, placing the work for thissoluton on your graph paper next toyour coordinate system containing thegraphical solution. Do the answersagree?

21. f(x) = 1x

+ 1x+ 5 , k = 9/14

22. f(x) = 1x

+ 1x! 2 , k = 8/15

23. f(x) = 1x! 1 !

1x+ 1 , k = 1/4

24. f(x) = 1x! 1 !

1x+ 2 , k = 1/6

25. f(x) = 1x! 2 + 1

x+ 2 , k = 4

26. f(x) = 1x! 3 + 1

x+ 2 , k = 5


27. 2x+ 1 + 4

x+ 2 = !3

28. 2x! 5 !

7x! 7 = 9

29. 3x+ 9 !

2x+ 7 = !3

30. 3x+ 9 !

6x+ 7 = 9

31. 2x+ 9 + 2

x+ 6 = !1

32. 5x! 6 !

8x! 7 = !1

33. 3x+ 3 + 6

x+ 2 = !2

34. 2x! 4 !

2x! 1 = 1

Section 7.7 Solving Rational Equations 725

Version: Fall 2007

For each of the equations in Exercises 35-40, perform each of the following tasks.

i. Follow the lead of Example 10 in thetext. Make one side of the equationequal to zero. Load the nonzero sideinto your calculator and draw its graph.

ii. Determine the vertical asymptotes ofby analyzing the equation and the re-sulting graph on your calculator. Usethe TABLE feature of your calculatorto determine any horizontal asymp-tote behavior.

iii. Use the zero finding utility in theCALC menu to determine the zero ofthe nonzero side of the resulting equa-tion.

iv. Set up a coordinate system on graphpaper. Label and scale each axis. Re-member to draw all lines with a ruler.Draw the graph of the nonzero sideof the equation. Draw the verticaland horizontal asymptotes and labelthem with their equations. Plot thex-intercept and label it with its coor-dinates.

v. Use an algebraic technique to deter-mine the solution of the equation andcompare it with the solution found bythe graphical analysis above.

35. x

x+ 1 + 8x2 ! 2x! 3 = 2

x! 3

36. x

x+ 4 !2x+ 1 = 12

x2 + 5x+ 4

37. x

x+ 1 !4

2x+ 1 = 2x! 12x2 + 3x+ 2

38. 2xx! 4 !

1x+ 1 = 4x+ 24

x2 ! 3x! 4

39. x

x! 2 + 3x+ 2 = 8

4! x2

40. x

x! 1 !4x+ 1 = x! 6

1! x2


41. x

3x! 9 !9x

= 1x! 3

42. 5xx+ 2 + 5

x! 5 = x+ 6x2 ! 3x! 10

43. 3xx+ 2 !

7x

= ! 12x+ 4

44. 4xx+ 6 !

4x+ 4 = x! 4

x2 + 10x+ 24

45. x

x! 5 + 94! x = x+ 5

x2 ! 9x+ 20

46. 6xx! 5 !

2x! 3 = x! 8

x2 ! 8x+ 15

47. 2xx! 4 + 5

2! x = x+ 8x2 ! 6x+ 8

48. x

x! 7 !8

5! x = x+ 7x2 ! 12x+ 35

49. ! x

2x+ 2 !6x

= ! 2x+ 1

50. 7xx+ 3 !

42! x = x+ 8

x2 + x! 6

51. 2xx+ 5 !

26! x = x! 2

x2 ! x! 30

52. 4xx+ 1 + 6

x+ 3 = x! 9x2 + 4x+ 3

53. x

x+ 7 !2x+ 5 = x+ 1

x2 + 12x+ 35

54. 5x6x+ 4 + 6

x= 1

3x+ 2

55. 2x3x+ 9 !

4x

= ! 2x+ 3


Version: Fall 2007

56. 7xx+ 1 !

4x+ 2 = x+ 6

x2 + 3x+ 2

57. x

2x! 8 + 8x

= 2x! 4

58. 3xx! 6 + 6

x! 6 = x+ 2x2 ! 12x+ 36

59. x

x+ 2 + 2x

= ! 52x+ 4

60. 4xx! 2 + 2

2! x = x+ 4x2 ! 4x+ 4

61. ! 2x3x! 9 !

3x

= ! 2x! 3

62. 2xx+ 1 !

2x

= 12x+ 2

63. x

x+ 1 + 5x

= 14x+ 4

64. 2xx! 4 !

8x! 7 = x+ 2

x2 ! 11x+ 28

65. ! 9x8x! 2 + 2

x= ! 2

4x! 1

66. 2xx! 3 !

44! x = x! 9

x2 ! 7x+ 12

67. 4xx+ 6 !

57 ! x = x! 5

x2 ! x! 42

68. x

x! 1 !4x

= 15x! 5

Section 7.7 Solving Rational Equations

Version: Fall 2007

7.7 Solutions

1. Load f(x) = (x ! 1)/(x + 2) into Y1 and y = 3 in Y2, as shown in (a). Use theintersect utility from the CALC menu to find the point of intersection, as shown in(b). Note: We had some di!culty with this at first, because the calculator warnedthat it could not find a point of intersection. We tried a second time, but this time wemade sure our “guess” was near the point of intersection (and certainly to the left ofthe vertical asymptote). This approach provided the solution in (b).

(a) (b)

Note that x = 1 makes the numerator of f(x) = (x!1)/(x+2) zero without making thedenominator zero. Hence, x = 1 is a zero of f and (1, 0) is an x-intercept of the graphof f . Secondly, the function f is reduced and x = !2 makes the denominator zero.Thus, x = !2 is a vertical asymptote. Tabular results indicate a horizontal asymptotey = 1.

(c) (d)

This is enough information to construct the graphs of f(x) = (x!1)/(x+ 2) and y = 3that follows. Note that we drop a dashed vertical line from the point of intersection tothe x-axis where we label the x-value.


Version: Fall 2007

x10

y10

x=!2

y=1(1,0)(1,0)

y=3(!3.5,3)(!3.5,3)

!3.5!3.5

To solve the equation algebraically, first multiply both sides of the equation by x+ 2.x! 1x+ 2 = 3

(x+ 2)!x! 1x+ 2

"= 3(x+ 2)

x! 1 = 3(x+ 2)

Expand, then solve for x.x! 1 = 3x+ 6x! 3x = 6 + 1!2x = 7x = !7/2

Note how the solution x = !7/2 agrees with the graphical solution found above.

3. Load f(x) = (x + 1)/(3 ! x) into Y1 and y = 2 in Y2, as shown in (a). Use theintersect utility from the CALC menu to find the point of intersection, as shown in(b). Note: We had some di!culty with this at first, because the calculator warnedthat it could not find a point of intersection. We tried a second time, but this time wemade sure our “guess” was near the point of intersection (and certainly to the left ofthe vertical asymptote). This approach provided the solution in (b).

(a) (b)

Note that x = !1 makes the numerator of f(x) = (x+ 1)/(3!x) zero without makingthe denominator zero. Hence, x = !1 is a zero of f and (!1, 0) is an x-intercept of thegraph of f . Secondly, the function f is reduced and x = 3 makes the denominator zero.


Version: Fall 2007

Thus, x = 3 is a vertical asymptote. Tabular results indicate a horizontal asymptotey = !1.

(c) (d)

This is enough information to construct the graphs of f(x) = (x+ 1)/(3!x) and y = 2that follows. Note that we drop a dashed vertical line from the point of intersection tothe x-axis where we label the x-value.

x10

y10

x=3

y=!1

(!1,0)(!1,0)y=2 (1.6667,2)(1.6667,2)

1.6667

To solve the equation algebraically, first multiply both sides of the equation by 3! x.x+ 13! x = 2

(3! x)!x+ 13! x

"= 2(3! x)

x+ 1 = 2(3! x)

Expand, then solve for x.

x+ 1 = 6! 2xx+ 2x = 6! 1

3x = 5x = 5/3

Note how the solution x = 5/3 agrees with the graphical solution found above.


Version: Fall 2007

5. Load f(x) = (2x + 3)/(x ! 1) into Y1 and y = !3 in Y2, as shown in (a). Usethe intersect utility from the CALC menu to find the point of intersection, as shownin (b). Note: We had some di!culty with this at first, because the calculator warnedthat it could not find a point of intersection. We tried a second time, but this time wemade sure our “guess” was near the point of intersection (and certainly to the left ofthe vertical asymptote). This approach provided the solution in (b).

(a) (b)

Note that x = !3/2 makes the numerator of f(x) = (2x + 3)/(x ! 1) zero withoutmaking the denominator zero. Hence, x = !3/2 is a zero of f and (!3/2, 0) is anx-intercept of the graph of f . Secondly, the function f is reduced and x = 1 makesthe denominator zero. Thus, x = 1 is a vertical asymptote. Tabular results indicate ahorizontal asymptote y = 2.

(c) (d)

This is enough information to construct the graphs of f(x) = (2x + 3)/(x ! 1) andy = !3 that follows. Note that we drop a dashed vertical line from the point ofintersection to the x-axis where we label the x-value.

x10

y10

x=1

y=2(!3/2,0)(!3/2,0)

y=!3 (0,!3)(0,!3)

0


Version: Fall 2007

To solve the equation algebraically, first multiply both sides of the equation by x! 1.2x+ 3x! 1 = !3

(x! 1)!2x+ 3x! 1

"= !3(x! 1)

2x+ 3 = !3(x! 1)

Expand, then solve for x.

2x+ 3 = !3x+ 32x+ 3x = 3! 3

5x = 0x = 0

Note how the solution x = 0 agrees with the graphical solution found above.

7. To solve 16x!92x!1 = 8, first clear fractions by multiplying both sides of the equation

by the LCD 2x! 1 to obtain

16x! 9 = 16x! 8

Then solve this linear equation. In this case, there are no solutions.

9. To solve 5x+84x+1 = !11, first clear fractions by multiplying both sides of the equation

by the LCD 4x+ 1 to obtain

5x+ 8 = !44x! 11

Then solve this linear equation to obtain the solution !1949 . Since this value does not

cause division by zero in the original equation, it is a valid solution.

11. To solve ! 35x7x+12 = !5, first clear fractions by multiplying both sides of the

equation by the LCD 7x+ 12 to obtain

!35x = !35x! 60

Then solve this linear equation. In this case, there are no solutions.

13. To solve 8x+2x!11 = 11, first clear fractions by multiplying both sides of the equation

by the LCD x! 11 to obtain

8x+ 2 = 11x! 121

Then solve this linear equation to obtain the solution 41. Since this value does notcause division by zero in the original equation, it is a valid solution.


Version: Fall 2007

15. The least common denominator (LCD) is 63, so first clear fractions by multiplyingboth sides of the equation by the LCD to obtain

9x+ 56 = !72

Then solve this linear equation to obtain the solution !1289 . Since this value does not

cause division by zero in the original equation, it is a valid solution.

17. The least common denominator (LCD) is x2, so first clear fractions by multiplyingboth sides of the equation by the LCD to obtain the quadratic equation

!57x = 27x2 ! 40

Then solve this quadratic equation either by factoring or by using the quadratic formulato obtain the solutions !8

3 and 59 . Since each of these values does not cause division by

zero in the original equation, they are both valid solutions.

19. The least common denominator (LCD) is x2, so first clear fractions by multiplyingboth sides of the equation by the LCD to obtain the quadratic equation

7x = 4x2 ! 3

Then solve this quadratic equation by using the quadratic formula to obtain the solu-tions 7+

"97

8 and 7!"

978 . Since each of these values does not cause division by zero in

the original equation, they are both valid solutions.

21. Load f(x) = 1/x + 1/(x + 5) into Y1 and y = 9/14 in Y2, as shown in (a). Usethe intersect utility from the CALC menu to find the points of intersection, as shownin (b) and (c).

(a) (b) (c)

We can combine terms of f .

f(x) = 1x

+ 1x+ 5 = x+ 5

x(x+ 5) + x

x(x+ 5) = 2x+ 5x(x+ 5)

Note that x = !5/2 makes the numerator of f(x) = (2x+ 5)/(x(x+ 5)) zero withoutmaking the denominator zero. Hence, x = !5/2 is a zero of f and (!5/2, 0) is anx-intercept of the graph of f . Secondly, the function f is reduced and x = !5 andx = 0 make the denominator zero. Thus, x = !5 and x = 0 are vertical asymptotes.Tabular results indicate a horizontal asymptote y = 0.


Version: Fall 2007

(d) (e)

This is enough information to construct the graphs of f(x) = (2x+ 5)/(x(x+ 5)) andy = 9/14 that follows. The x-value of each point of intersection is a solution of theequation f(x) = 9/14. Note: In this instance, we avoided the usual dashed verticallines because we thought the final picture would be too crowded.

x10

y10

x=0x=!5

y=0(!2.5,0)(!2.5,0)

y=9/14(!3.8889,9/14)(!3.8889,9/14) (2,9/14)(2,9/14)

To solve the equation algebraically, first multiply both sides of the equation by 14x(x+5).

2x+ 5x(x+ 5) = 9

14

14x(x+ 5)! 2x+ 5x(x+ 5)

"= 9

1414x(x+ 5)

14(2x+ 5) = 9x(x+ 5)

Expand, make one side zero.

28x+ 70 = 9x2 + 45x0 = 9x2 + 17x! 70

Let’s use the quadratic formula.

x = !17 ±#

172 ! 4(9)(!70)2(9) = !17 ±

"2809

18 = !17 ± 5318

This gives us two answers,

x = !17 ! 5318 = !35

9 and x = !17 + 5318 = 2.


Version: Fall 2007

Note how these solutions agree with the graphical solutions found above.

23. Load f(x) = 1/(x!1)!1/(x+1) into Y1 and y = 1/4 in Y2, as shown in (a). Usethe intersect utility from the CALC menu to find the points of intersection, as shownin (b) and (c).

(a) (b) (c)


f(x) = 1x! 1 !

1x+ 1 = x+ 1

(x! 1)(x+ 1) !x! 1

(x! 1)(x+ 1) = 2(x! 1)(x+ 1)

There is no value of x that will make the numerator zero. Hence, the graph of f hasno x-intercept. Secondly, the function f is reduced and x = !1 and x = 1 make thedenominator zero. Thus, x = !1 and x = 1 are vertical asymptotes. Tabular resultsindicate a horizontal asymptote y = 0.

(d) (e)

This is enough information to construct the graphs of f(x) = 2/((x ! 1)(x + 1)) andy = 1/4 that follows. The x-value of each point of intersection is a solution of theequation f(x) = 1/4. Note: In this instance, we avoided the usual dashed vertical linesbecause we thought the final picture would be too crowded.

x10

y10

x=!1 x=1

y=0 y=1/4(!3,1/4)(!3,1/4) (3,1/4)(3,1/4)


Version: Fall 2007

To solve the equation algebraically, first multiply both sides of the equation by 4(x !1)(x+ 1).

2(x! 1)(x+ 1) = 1

4

4(x! 1)(x+ 1)! 2

(x! 1)(x+ 1)

"= 1

44(x! 1)(x+ 1)

8 = x2 ! 19 = x2

x = ±3


25. Load f(x) = 1/(x! 2) + 1/(x+ 2) into Y1 and y = 4 in Y2, as shown in (a). Usethe intersect utility from the CALC menu to find the points of intersection, as shownin (b) and (c).

(a) (b) (c)


f(x) = 1x! 2 + 1

x+ 2 = x+ 2(x! 2)(x+ 2) + x! 2

(x! 2)(x+ 2) = 2x(x! 2)(x+ 2)

Note that x = 0 makes the numerator equal to zero without making the denominatorzero. Hence, x = 0 is a zero of f and (0, 0) is an x-intercept of the graph of f . Secondly,the function f is reduced and x = !2 and x = 2 make the denominator zero. Thus,x = !2 and x = 2 are vertical asymptotes. Tabular results indicate a horizontalasymptote y = 0.

(d) (e)

This is enough information to construct the graphs of f(x) = 2x/((x! 2)(x+ 2)) andy = 4 that follows. The x-value of each point of intersection is a solution of the equation


Version: Fall 2007

f(x) = 4. Note: In this instance, we avoided the usual dashed vertical lines becausewe thought the final picture would be too crowded.

x10

y10

x=!2 x=2

y=0(0,0)(0,0)

y=4(!1.765564,4)(!1.765564,4) (2.2655644,4)(2.2655644,4)

To solve the equation algebraically, first multiply both sides of the equation by (x !2)(x+ 2).

2x(x! 2)(x+ 2) = 4

(x! 2)(x+ 2)! 2x

(x! 2)(x+ 2)

"= 4(x! 2)(x+ 2)

2x = 4(x2 ! 4)2x = 4x2 ! 16

Make one side zero and divide both side of the resulting equation by 2.

0 = 4x2 ! 2x! 160 = 2x2 ! x! 8


x = !(!1)±#

(!1)2 ! 4(2)(!8)2(2) = 1±

"65

4

If you approximate these with your calculator, you will see that they match the ap-proximations found graphically above.

(f) (g)



Version: Fall 2007

27. The least common denominator (LCD) is (x + 1)(x + 2), so first clear fractionsby multiplying both sides of the equation by the LCD to obtain the equation

2(x+ 2) + 4(x+ 1) = !3(x+ 1)(x+ 2)

Then simplify this quadratic equation to standard form:

!3x2 ! 15x! 14 = 0

Finally, solve this quadratic equation by using the quadratic formula to obtain thesolutions !15+

"57

6 and !15!"

576 . Since each of these values does not cause division by



3(x+ 7)! 2(x+ 9) = !3(x+ 9)(x+ 7)


!3x2 ! 49x! 192 = 0


"97

6 and !49!"




2(x+ 6) + 2(x+ 9) = !1(x+ 9)(x+ 6)


!x2 ! 19x! 84 = 0

Finally, solve this quadratic equation either by factoring or by using the quadraticformula to obtain the solutions !7 and !12. Since each of these values does not causedivision by zero in the original equation, they are both valid solutions.


3(x+ 2) + 6(x+ 3) = !2(x+ 3)(x+ 2)


!2x2 ! 19x! 36 = 0


"73

4 and !19!"




Version: Fall 2007

35. Setx

x+ 1 + 8x2 ! 2x! 3 !

2x! 3 = 0,

and load the left-hand side into Y1, as shown in (a). Use the zero finding utility in theCALC menu to determine the zero, as shown in (b).

(a) (b)

The equation would indicate vertical asymptotes at x = !1 and x = 3, but the image in(b) would indicate that some sort of cancellation takes place, leaving only one verticalasymptote at x = !1. The following tables indicate the presence of a horizontalasymptote y = 1.

(c) (d)

Copy the image onto your homework as follows.

x10

y10

x=!1

y=1(2,0)(2,0)

To solve the equation algebraically, multiply both sides of the equation by the commondenominator (x+ 1)(x! 3).


Version: Fall 2007

x

x+ 1 + 8x2 ! 2x! 3 = 2

x! 3

(x+ 1)(x! 3)!x

x+ 1 + 8(x+ 1)(x! 3)

"=! 2x! 3

"(x+ 1)(x! 3)

x(x! 3) + 8 = 2(x+ 1)

Expand, then make one side zero and factor.

x2 ! 3x+ 8 = 2x+ 2x2 ! 5x+ 6 = 0

(x! 3)(x! 2) = 0

This would indicate two solutions. However, note that x = 3 is an extraerroeous answer,as it makes some rational expressions in the original equation undefined. Hence, x = 2is the only solution. Note that this agrees with the graphical solution above.

37. Setx

x+ 1 !4

2x+ 1 !2x! 1

2x2 + 3x+ 1 = 0


(a) (b)

The equation would indicate vertical asymptotes at x = !1 and x = !1/2, but theimage in (b) would indicate that some sort of cancellation takes place, leaving only onevertical asymptote at x = !1. The following tables indicate the presence of a horizontalasymptote y = 1.

(c) (d)


Version: Fall 2007


x10

y10

x=!1

y=1(3,0)(3,0)

To solve the equation algebraically, multiply both sides of the equation by the commondenominator (x+ 1)(2x+ 1).

x

x+ 1 !4

2x+ 1 = 2x! 12x2 + 3x+ 1

(x+ 1)(2x+ 1)!x

x+ 1 !4

2x+ 1

"=! 2x! 1

(2x+ 1)(x+ 1)

"(x+ 1)(2x+ 1)

x(2x+ 1)! 4(x+ 1) = 2x! 1Expand, simplify, then make one side zero and factor.

2x2 + x! 4x! 4 = 2x! 12x2 ! 3x! 4 = 2x! 12x2 ! 5x! 3 = 0

(2x+ 1)(x! 3) = 0This would indicate two solutions. However, note that x = !1/2 is an extraerroeousanswer, as it makes some rational expressions in the original equation undefined. Hence,x = 3 is the only solution. Note that this agrees with the graphical solution above.

39. Setx

x! 2 + 3x+ 2 !

84! x2 = 0


(a) (b) (c)


Version: Fall 2007

The equation and graph indicate vertical asymptotes at x = !2 and x = 2. Thefollowing tables indicate the presence of a horizontal asymptote y = 1.

(d) (e)


x10

y10

x=!2 x=2

y=1((!5!

"17)/2,0)((!5!"

17)/2,0) ((!5+"

17)/2,0)((!5+"

17)/2,0)

First, make a sign change, negating two parts of the last rational expression on theright-hand side of the equation.

x

x! 2 + 3x+ 2 = !8

x2 ! 4 (1)

To solve the equation algebraically, multiply both sides of the equation by the commondenominator (x+ 2)(x! 2).

x

x! 2 + 3x+ 2 = !8

(x+ 2)(x! 2)

(x+ 2)(x! 2)!x

x! 2 + 3x+ 2

"=!

!8(x+ 2)(x! 2)

"(x+ 2)(x! 2)

x(x+ 2) + 3(x! 2) = !8

Expand, simplify, then make one side zero and factor.

x2 + 2x+ 3x! 6 = !8x2 + 5x! 6 = !8x2 ! 5x+ 2 = 0


Version: Fall 2007


x = !5±#

52 ! 4(1)(2)2(1) = !5±

"17

2

To compare these exact solutions to the approximate solutions found by graphicalanalysis above, use your calculator to find decimal approximations, as shown in (f) and(g) below.

(f) (g)

41. The least common denominator (LCD) is x(3x! 9), so multiply both sides of theequation by the LCD to obtain

x2 ! 9(3x! 9) = 3x

Then solve this quadratic equation either by factoring or by using the quadratic formulato obtain the solutions 27 and 3. However, 3 causes division by zero when substitutedfor x in the original equation. Therefore, 27 is the only valid solution.

43. The least common denominator (LCD) is x(2x+ 4), so multiply both sides of theequation by the LCD to obtain

6x2 ! 7(2x+ 4) = !x

Then solve this quadratic equation either by factoring or by using the quadratic formulato obtain the solutions 7

2 and !43 . Since each of these values does not cause division

by zero in the original equation, they are both valid solutions.

45. The least common denominator (LCD) is (x!5)(x!4). However, the denominatorof the second term is 4!x, so first rewrite the left side by negating the numerator anddenominator of the second term:

x

x! 5 !9x! 4 = x+ 5

x2 ! 9x+ 20Then clear fractions by multiplying both sides of the equation by the LCD to obtainthe equation

x(x! 4)! 9(x! 5) = x+ 5



Version: Fall 2007

x2 ! 14x+ 40 = 0

Finally, solve this quadratic equation either by factoring or by using the quadraticformula to obtain the solutions 10 and 4. However, 4 causes division by zero whensubstituted for x in the original equation. Therefore, 10 is the only valid solution.

47. The least common denominator (LCD) is (x!4)(x!2). However, the denominatorof the second term is 2!x, so first rewrite the left side by negating the numerator anddenominator of the second term:

2xx! 4 !

5x! 2 = x+ 8

x2 ! 6x+ 8Then clear fractions by multiplying both sides of the equation by the LCD to obtainthe equation

2x(x! 2)! 5(x! 4) = x+ 8


2x2 ! 10x+ 12 = 0

Finally, solve this quadratic equation either by factoring or by using the quadraticformula to obtain the solutions 3 and 2. However, 2 causes division by zero whensubstituted for x in the original equation. Therefore, 3 is the only valid solution.


!x2 ! 6(2x+ 2) = !4x

Then solve this quadratic equation either by factoring or by using the quadratic formulato obtain the solutions !6 and !2. Since each of these values does not cause divisionby zero in the original equation, they are both valid solutions.

51. The least common denominator (LCD) is (x+5)(x!6). However, the denominatorof the second term is 6!x, so first rewrite the left side by negating the numerator anddenominator of the second term:

2xx+ 5 + 2

x! 6 = x! 2x2 ! x! 30

Then clear fractions by multiplying both sides of the equation by the LCD to obtainthe equation

2x(x! 6) + 2(x+ 5) = x! 2


2x2 ! 11x+ 12 = 0


Version: Fall 2007

Finally, solve this quadratic equation either by factoring or by using the quadraticformula to obtain the solutions 4 and 3

2 . Since each of these values does not causedivision by zero in the original equation, they are both valid solutions.


x(x+ 5)! 2(x+ 7) = x+ 1


x2 + 2x! 15 = 0

Finally, solve this quadratic equation either by factoring or by using the quadraticformula to obtain the solutions 3 and !5. However, !5 causes division by zero whensubstituted for x in the original equation. Therefore, 3 is the only valid solution.


2x2 ! 4(3x+ 9) = !6x

Then solve this quadratic equation either by factoring or by using the quadratic formulato obtain the solutions 6 and !3. However, !3 causes division by zero when substitutedfor x in the original equation. Therefore, 6 is the only valid solution.


x2 + 8(2x! 8) = 4x

Then solve this quadratic equation either by factoring or by using the quadratic formulato obtain the solutions 4 and !16. However, 4 causes division by zero when substitutedfor x in the original equation. Therefore, !16 is the only valid solution.


2x2 + 2(2x+ 4) = !5x

Then solve this quadratic equation by using the quadratic formula to obtain the solu-tions

!9 +"

174 and !9!

"17

4Since each of these values does not cause division by zero in the original equation, theyare both valid solutions.


Version: Fall 2007


!2x2 ! 3(3x! 9) = !6x

Then solve this quadratic equation either by factoring or by using the quadratic formulato obtain the solutions !9

2 and 3. However, 3 causes division by zero when substitutedfor x in the original equation. Therefore, !9

2 is the only valid solution.


4x2 + 5(4x+ 4) = x

Then solve this quadratic equation by using the quadratic formula to obtain the solu-tions

!19 +"

418 and !19!

"41

8Since each of these values does not cause division by zero in the original equation, theyare both valid solutions.


!9x2 + 2(8x! 2) = !4x

Then solve this quadratic equation either by factoring or by using the quadratic formulato obtain the solutions 2

9 and 2. Since each of these values does not cause division byzero in the original equation, they are both valid solutions.

67. The least common denominator (LCD) is (x+6)(x!7). However, the denominatorof the second term is 7!x, so first rewrite the left side by negating the numerator anddenominator of the second term:

4xx+ 6 + 5

x! 7 = x! 5x2 ! x! 42

Then clear fractions by multiplying both sides of the equation by the LCD to obtainthe equation

4x(x! 7) + 5(x+ 6) = x! 5


4x2 ! 24x+ 35 = 0

Finally, solve this quadratic equation either by factoring or by using the quadraticformula to obtain the solutions 7

2 and 52 . Since each of these values does not cause

division by zero in the original equation, they are both valid solutions.

Section 7.8 Applications of Rational Functions 743

Version: Fall 2007

7.8 Exercises

1. The sum of the reciprocals of twoconsecutive odd integers is !16

63 . Findthe two numbers.

2. The sum of the reciprocals of twoconsecutive odd integers is 28

195 . Find thetwo numbers.

3. The sum of the reciprocals of twoconsecutive integers is!19

90 . Find the twonumbers.

4. The sum of a number and its recip-rocal is 41

20 . Find the number(s).

5. The sum of the reciprocals of twoconsecutive even integers is 5


6. The sum of the reciprocals of twoconsecutive integers is 19

90 . Find the twonumbers.

7. The sum of a number and twice itsreciprocal is 9


8. The sum of a number and its recip-rocal is 5


9. The sum of the reciprocals of twoconsecutive even integers is 11


10. The sum of a number and twice itsreciprocal is 17


11. The sum of the reciprocals of twonumbers is 15/8, and the second numberis 2 larger than the first. Find the twonumbers.


12. The sum of the reciprocals of twonumbers is 16/15, and the second num-ber is 1 larger than the first. Find thetwo numbers.

13. Moira can paddle her kayak at aspeed of 2 mph in still water. She pad-dles 3 miles upstream against the cur-rent and then returns to the starting lo-cation. The total time of the trip is 9hours. What is the speed (in mph) ofthe current? Round your answer to thenearest hundredth.

14. Boris is kayaking in a river with a 6mph current. Suppose that he can kayak4 miles upstream in the same amount oftime as it takes him to kayak 9 milesdownstream. Find the speed (mph) ofBoris’s kayak in still water.

15. Jacob can paddle his kayak at aspeed of 6 mph in still water. He pad-dles 5 miles upstream against the cur-rent and then returns to the starting lo-cation. The total time of the trip is 5hours. What is the speed (in mph) ofthe current? Round your answer to thenearest hundredth.

16. Boris can paddle his kayak at a speedof 6 mph in still water. If he can paddle5 miles upstream in the same amount oftime as it takes his to paddle 9 milesdownstream, what is the speed of thecurrent?


Version: Fall 2007

17. Jacob is canoeing in a river with a5 mph current. Suppose that he can ca-noe 4 miles upstream in the same amountof time as it takes him to canoe 8 milesdownstream. Find the speed (mph) ofJacob’s canoe in still water.

18. The speed of a freight train is 16mph slower than the speed of a passengertrain. The passenger train travels 518miles in the same time that the freighttrain travels 406 miles. Find the speedof the freight train.


20. Emily can paddle her canoe at aspeed of 2 mph in still water. She pad-dles 5 miles upstream against the cur-rent and then returns to the starting lo-cation. The total time of the trip is 6hours. What is the speed (in mph) ofthe current? Round your answer to thenearest hundredth.

21. Jacob is canoeing in a river with a2 mph current. Suppose that he can ca-noe 2 miles upstream in the same amountof time as it takes him to canoe 5 milesdownstream. Find the speed (mph) ofJacob’s canoe in still water.

22. Moira can paddle her kayak at aspeed of 2 mph in still water. If shecan paddle 4 miles upstream in the sameamount of time as it takes her to paddle8 miles downstream, what is the speed ofthe current?

23. Boris can paddle his kayak at a speedof 6 mph in still water. If he can paddle5 miles upstream in the same amount oftime as it takes his to paddle 10 milesdownstream, what is the speed of thecurrent?


25. It takes Jean 15 hours longer tocomplete an inventory report than it takesSanjay. If they work together, it takesthem 10 hours. How many hours wouldit take Sanjay if he worked alone?

26. Jean can paint a room in 5 hours.It takes Amelie 10 hours to paint thesame room. How many hours will it takeif they work together?

27. It takes Amelie 18 hours longer tocomplete an inventory report than it takesJean. If they work together, it takesthem 12 hours. How many hours wouldit take Jean if she worked alone?

28. Sanjay can paint a room in 5 hours.It takes Amelie 9 hours to paint the sameroom. How many hours will it take ifthey work together?

29. It takes Ricardo 12 hours longerto complete an inventory report than ittakes Sanjay. If they work together, ittakes them 8 hours. How many hourswould it take Sanjay if he worked alone?

Section 7.8 Applications of Rational Functions 745

Version: Fall 2007

30. It takes Ricardo 8 hours longer tocomplete an inventory report than it takesAmelie. If they work together, it takesthem 3 hours. How many hours would ittake Amelie if she worked alone?

31. Jean can paint a room in 4 hours.It takes Sanjay 7 hours to paint the sameroom. How many hours will it take ifthey work together?

32. Amelie can paint a room in 5 hours.It takes Sanjay 9 hours to paint the sameroom. How many hours will it take ifthey work together?


Version: Fall 2007

7.8 Solutions

1. Start with the equation1x

+ 1x+ 2 = !16

63Multiply both sides by the LCD 63x(x+ 2):

63(x+ 2) + 63x = !16x(x+ 2)

Simplify to obtain

126x+ 126 = !16x2 ! 32x

This quadratic equation has one integer solution x = !9, and the second number is!9 + 2 = !7.


+ 1x+ 1 = !19


90(x+ 1) + 90x = !19x(x+ 1)

Simplify to obtain

180x+ 90 = !19x2 ! 19x

This quadratic equation has one integer solution x = !10, and the second number is!10 + 1 = !9.


+ 1x+ 2 = 5


12(x+ 2) + 12x = 5x(x+ 2)

Simplify to obtain

24x+ 24 = 5x2 + 10x

This quadratic equation has one integer solution x = 4, and the second number is4 + 2 = 6.

Section 7.8 Applications of Rational Functions

Version: Fall 2007

7. Start with the equation

x+ 2!1x

"= 9

2

Multiply both sides by the LCD 2x:

2x2 + 4 = 9x

Solve this quadratic equation to get x = 12 , 4.


+ 1x+ 2 = 11


60(x+ 2) + 60x = 11x(x+ 2)

Simplify to obtain

120x+ 120 = 11x2 + 22x

This quadratic equation has one integer solution x = 10, and the second number is10 + 2 = 12.


+ 1x+ 2 = 15


8(x+ 2) + 8x = 15x(x+ 2)

Simplify to obtain

16x+ 16 = 15x2 + 30x

Solve this quadratic equation to get x = 23 ,!85 . For x = 2

3 , the other number isx + 2 = 2

3 + 2 = 83 ; and, for x = !8

5 , the other number is x + 2 = !85 + 2 = 2

5 . Thus,the possible pairs of numbers are {2

3 ,83} and {!8

5 ,25}.


Version: Fall 2007

13. Let t1 represent the time it takes for the upstream part of the trip, and t2 representthe time it takes for the downstream part of the trip. Thus,

t1 + t2 = 9 hours

Also, let c represent the speed of the current. Then the actual speed of the boat goingupstream is 2! c, and the speed downstream is 2 + c. Using the relationship

distance = rate · time,

this leads to the following two equations:

Trip upstream: 3 = (2! c)t1Trip downstream: 3 = (2 + c)t2

Solving for the time variable in each equation, it follows that

t1 = 32! c and t2 = 3

2 + cTherefore,

9 = t1 + t2 = 32! c + 3

2 + cNow solve this rational equation for c:

9 = 32! c + 3

2 + c =" 9(2! c)(2 + c) = 3(2 + c) + 3(2! c)

=" 36! 9c2 = 12=" 9c2 = 24

=" c = ±#

83

Discarding the negative answer, the speed of the current is # 1.63 mph.

15. Let t1 represent the time it takes for the upstream part of the trip, and t2 representthe time it takes for the downstream part of the trip. Thus,

t1 + t2 = 5 hours

Also, let c represent the speed of the current. Then the actual speed of the boat goingupstream is 6! c, and the speed downstream is 6 + c. Using the relationship



Trip upstream: 5 = (6! c)t1Trip downstream: 5 = (6 + c)t2


Version: Fall 2007

Solving for the time variable in each equation, it follows that

t1 = 56! c and t2 = 5

6 + cTherefore,

5 = t1 + t2 = 56! c + 5

6 + cNow solve this rational equation for c:

5 = 56! c + 5

6 + c =" 5(6! c)(6 + c) = 5(6 + c) + 5(6! c)

=" 180! 5c2 = 60=" 5c2 = 120

=" c = ±$

24

Discarding the negative answer, the speed of the current is # 4.90 mph.

17. Let t represent the time it takes for each part of the trip, and let r represent thespeed of the boat in still water. Then the actual speed of the boat going downstreamis r + 5, and the speed upstream is r ! 5. Using the relationship



Trip downstream: 8 = (r + 5)tTrip upstream: 4 = (r ! 5)t

Solving for t in each equation, it follows that8r + 5 = t = 4

r ! 5Now solve this rational equation for r:

8r + 5 = 4

r ! 5 =" 8(r ! 5) = 4(r + 5)

=" 4r = 60=" r = 15 mph


Version: Fall 2007

19. Let t represent the time it takes for each trip, and let r represent the speed of thefreight train. Then the speed of the passenger train is r + 20. Using the relationship



Freight train: 280 = rtPassenger train: 440 = (r + 20)t

Solving for t in each equation, it follows that280r

= t = 440r + 20

Now solve this rational equation for r:280r

= 440r + 20 =" 280(r + 20) = 440r

=" 5600 = 160r=" r = 35 mph

21. Let t represent the time it takes for each part of the trip, and let r represent thespeed of the boat in still water. Then the actual speed of the boat going downstreamis r + 2, and the speed upstream is r ! 2. Using the relationship



Trip downstream: 5 = (r + 2)tTrip upstream: 2 = (r ! 2)t

Solving for t in each equation, it follows that5r + 2 = t = 2

r ! 2Now solve this rational equation for r:

5r + 2 = 2

r ! 2 =" 5(r ! 2) = 2(r + 2)

=" 3r = 14

=" r = 143 mph


Version: Fall 2007

23. Let t represent the time it takes for each part of the trip, and let c represent thespeed of the current. Then the actual speed of the boat going downstream is 6 + c, andthe speed upstream is 6! c. Using the relationship



Trip downstream: 10 = (6 + c)tTrip upstream: 5 = (6! c)t

Solving for t in each equation, it follows that10

6 + c = t = 56! c

Now solve this rational equation for r:10

6 + c = 56! c =" 10(6! c) = 5(6 + c)

=" 30 = 15c=" c = 2 mph

25. Let t represent the unknown time, r represent Jean’s rate (reports/hour), and srepresent Sanjay’s rate. Using the relationship

work = rate · time,

this leads to the following three equations:

Jean: 1 = r(t+ 15)Sanjay: 1 = st

together: 1 = (r + s) 10

Solve for r and s in the first two equations, and then plug those values into the thirdequation to obtain

1 =! 1t+ 15 + 1

t

"10

Now solve this rational equation for t:

1 =! 1t+ 15 + 1

t

"10 =" t(t+ 15) = 10t+ 10(t+ 15)

=" t2 + 15t = 20t+ 150=" t2 ! 5t! 150 = 0=" (t! 15)(t+ 10) = 0=" t = 15 hours


Version: Fall 2007

27. Let t represent the unknown time, r represent Amelie’s rate (reports/hour), ands represent Jean’s rate. Using the relationship



Amelie: 1 = r(t+ 18)Jean: 1 = st



1 =! 1t+ 18 + 1

t

"12


1 =! 1t+ 18 + 1

t

"12 =" t(t+ 18) = 12t+ 12(t+ 18)

=" t2 + 18t = 24t+ 216=" t2 ! 6t! 216 = 0=" (t! 18)(t+ 12) = 0=" t = 18 hours

29. Let t represent the unknown time, r represent Ricardo’s rate (reports/hour), ands represent Sanjay’s rate. Using the relationship



Ricardo: 1 = r(t+ 12)Sanjay: 1 = st



1 =! 1t+ 12 + 1

t

"8


Version: Fall 2007


1 =! 1t+ 12 + 1

t

"8 =" t(t+ 12) = 8t+ 8(t+ 12)

=" t2 + 12t = 16t+ 96=" t2 ! 4t! 96 = 0=" (t! 12)(t+ 8) = 0=" t = 12 hours

31. Let t represent the unknown time, r represent Jean’s rate (rooms/hour), and srepresent Sanjay’s rate. Using the relationship



Jean: 1 = 4rSanjay: 1 = 7s

together: 1 = (r + s) t


1 =!1

4 + 17

"t


1 =!1

4 + 17

"t =" 1 =

!1128

"t =" t = 28

11 hours

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7.1 Exercises